HSC 2015 MX2 Integration Marathon (archive) (1 Viewer)

omegadot · Oct 22, 2015

Re: MX2 2015 Integration Marathon

juantheron said:
$(1)\; If $f(x) = x^3+3x+4\;,$ Then value of $\int_{-1}^{1}f(x)dx+\int_{0}^{4}f^{-1}(x)dx =$

$A small \textbf{Lemma} to begin with\\If $f^{-1}(x)$ is invertible on the interval $a \leqslant x \leqslant b$, then\\$\int^b_a f^{-1}(x) \, dx = b \cdot f^{-1}(b) - a \cdot f^{-1}(a) - \int^{f^{-1}(b)}_{f^{-1}(a)} f(x) \, dx.\\$\textbf{Proof}$\\$Let $y = f^{-1}(x) \Rightarrow x = f(y),$ as $f^{-1}$ is invertible. So $dx = f'(y) dy$ and we have$\\\begin{align*}\int^b_a f^{-1}(x) \, dx &= \int^{f^{-1}(b)}_{f^{-1}(a)} y f'(y) \, dy\\&= y f(y) \Big{|}^{f^{-1}(b)}_{f^{-1}(a)} - \int^{f^{-1}(b)}_{f^{-1}(a)} f(y) \, dy \quad \mbox{(by parts)}\\&= b \cdot f^{-1}(b) - a \cdot f^{-1}(a) - \int^{f^{-1}(b)}_{f^{-1}(a)} f(x) \, dx\\\end{align*}$\\and completes the proof. (to be continued...)$

omegadot · Oct 22, 2015

Re: MX2 2015 Integration Marathon

juantheron said:
$$(1)\; If $f(x) = x^3+3x+4\;,$ Then value of $\int_{-1}^{1}f(x)dx+\int_{0}^{4}f^{-1}(x)dx =$[/QUOTE]$Now for the function $f$, as $f'(x) = 3x^2 + 3 > 0$ for all $x$ it is an increasing function on its domain and is therefore one-to-one and hence possesses an inverse. We begin by noting that$\\f(-1) = 0 \Rightarrow f^{-1}(0) = -1 \quad \mbox{and} \quad f(0) = 4 \Rightarrow f^{-1}(4) = 0.$\\Now applying the lemma to the integral containing the inverse we have$\\\int^4_0 f^{-1}(x) \, dx = 4 \cdot f^{-1}(4) - 0 \cdot f^{-1}(0) - \int^{f^{-1}(4)}_{f^{-1}(0)} f(x) \, dx=-\int^0_{-1} f(x) \, dx\\$So$\\\begin{align*}\int^1_{-1} f(x) \, dx + \int^4_0 f^{-1}(x) \, dx &= \int^{1}_{-1} f(x) \, dx - \int^0_{-1} f(x) \, dx\\&= \int^0_{-1} f(x) \, dx + \int^1_0 f(x) \, dx - \int^0_{-1} f(x) \, dx\\&= \int^1_0 f(x) \, dx\\&= \int^1_0 (x^3 + 3x + 4) dx\\&= \left [\frac{x^4}{4} + \frac{3x^2}{2} + 4x \right ]^1_0\\&=\frac{23}{4}\end{align*}$

omegadot · Oct 22, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
$\int {\tan^2{(x-a)}\tan^2{(x-b)}}dx$

$$Using$\\\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B},$ after rearranging we have$\\\tan A + \tan B = \frac{\tan A - \tan B}{\tan (A - B)} - 1.\\$Setting $ A = x - a$ and $B = x - b$ we have$\\\begin{align*}\tan (x - a) \tan (x - b) &= \frac{\tan(x - a) - \tan (x - b)}{\tan (b - a)} - 1\\&= \alpha [\tan(x - a) - \tan (x - b)] - 1,\end{align*}\\$where we have set $\alpha = 1/\tan(b - a) = \cot (b - a).$

$$So$\\\begin{align*}\tan^2(x - a) \tan^2 (x - b) &= [\alpha(\tan(x - a) - \tan (x - b)) - 1]^2\\&=\alpha^2[\tan^2(x - a) - 2 \tan (x - a) \tan (x - b) + \tan^2 (x - b)] - 2\alpha [\tan(x - a) - \tan (x - b)] + 1\\&=\alpha^2 [\sec^2 (x - a) -1 -2 \{\alpha [\tan(x - a) - \tan (x - b)] - 1\} + \sec^2 (x - b) - 1] - 2\alpha [\tan(x - a) - \tan (x - b)] + 1\\&=\alpha^2 [\sec^2 (x - a) + \sec^2 (x - b)] - 2\alpha (\alpha^2 + 1) [\tan(x - a) - \tan (x - b)] + 1.\\\end{align*}$\\$

$$Integrating we have$\\\begin{align*}\int \tan^2 (x - a) \tan^2 (x - b) dx &= \int x dx - 2 \alpha (\alpha^2 + 1) \int [\tan(x - a) - \tan (x - b)] dx + \alpha^2 \int [\sec^2 (x - a) + \sec^2 (x - b)] dx\\&=x - 2 \alpha (\alpha^2 + 1) \left (\ln |\sec (x - a)| - \ln |\sec(x - b)| \right ) + \alpha^2 [\tan(x - a) + \tan (x - b)] + \cal{C}\end{align*}\\$Noting that $\alpha^2 + 1 = 1 + \cot^2 (b - a) = \csc^2 (b - a)$ we finally have$\\\begin{align*}\int \tan^2 (x - a) \tan^2 (x - b) dx = x - 2 \cot (b - a) \csc^2(b - a) \ln \left |\frac{\cos (x - b)}{\cos (x - a)} \right | + \cot^2 (b - a) [\tan(x - a) + \tan (x - b)] + \cal{C}.\end{align*}$

omegadot · Oct 23, 2015

Re: MX2 2015 Integration Marathon

$$(a)$ \int \frac{\sin^{-1} x}{x^2} dx \qquad $(b)$ \int \left (\sin^{-1} x \right )^3 dx.$

turntaker · Oct 23, 2015

Re: MX2 2015 Integration Marathon

omegadot said:
$$(a)$ \int \frac{\sin^{-1} x}{x^2} dx \qquad $(b)$ \int \left (\sin^{-1} x \right )^3 dx.$

is A) integration by parts?

InteGrand · Oct 23, 2015

Re: MX2 2015 Integration Marathon

turntaker said:
is A) integration by parts?

$Yes:$

$Let $I=\int \frac{\sin ^{-1}x}{x^2}\text{ d}x$. By integration by parts, we have$

$I = -\frac{1}{x}\sin ^{-1}x - \int -\frac{1}{x}\times \frac{1}{\sqrt{1-x^2}}\text{ d}x$

$$= -\frac{1}{x}\sin ^{-1}x + \underbrace{\int \frac{1}{x\sqrt{1-x^2}}\text{ d}x}_{J}$.$

$To find $J$, let $\theta = \sin ^{-1}x$, so $\mathrm{d}\theta = \frac{1}{\sqrt{1-x^2}}\text{ d}x$ and $x = \sin \theta$. Substituting, we have$

$J = \int \frac{\mathrm{d}\theta}{\sin \theta}$

$= \int \csc \theta \text{ d}\theta$

$= -\ln \left |\csc \theta + \cot \theta \right |+\mathcal{C}$

$= -\ln \left |\frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta} \right |+\mathcal{C}$

$$= -\ln \left |\frac{1}{\sin \theta} + \frac{\sqrt{1-\sin^2 \theta}}{\sin \theta} \right |+\mathcal{C}$ \Big{(}as $\cos \theta \geq 0$, as $\theta = \sin^{-1}x \in \left[-\frac{\pi}{2},\frac{\pi}{2} \right]$\Big{)}$

$$= -\ln \left |\frac{1}{x} + \frac{\sqrt{1-x^2}}{x} \right |+\mathcal{C}$.$

$Therefore, $I= -\frac{1}{x}\sin ^{-1}x -\ln \left |\frac{1}{x} + \frac{\sqrt{1-x^2}}{x} \right |+\mathcal{C}$.$

omegadot · Oct 23, 2015

Re: MX2 2015 Integration Marathon

turntaker said:
is A) integration by parts?

$Another way to get there is as follows. Let $I = \int \frac{\sin^{-1} x}{x^2} dx$ and let $y = \sin^{-1} x \,\, \Rightarrow \,\, x = \sin y.$ So $dx = \cos y \, dy$ and we have$

$\begin{align*}I &= \int \frac{y}{\sin^2 y} \cos y \, dy\\&= \int y \cot y \csc y \, dy\\&= -y \csc y + \int \csc y \, dy \quad (\mbox{by parts})\\&= -y \csc y - \ln \left |\csc y + \cot y \right | + \cal{C}\end{align*}$

$$Then following InteGrand's solution, we have$\\\begin{align*}I &= -\frac{y}{\sin y} - \ln \left |\frac{1}{\sin y} + \frac{\cos y}{\sin y} \right | + \cal{C}\\&= -\frac{y}{\sin y} - \ln \left |\frac{1}{\sin y} + \frac{\sqrt{1 - \sin^2y}}{\sin y} \right | + \cal{C}\\&= -\frac{\sin^{-1} x}{x} - \ln \left |\frac{1 + \sqrt{1 - x^2}}{x} \right | + \cal{C}\end{align*}$

omegadot · Oct 24, 2015

Re: MX2 2015 Integration Marathon

juantheron said:
$(3)\; Let $f(k) = \int_{-1}^{1}\frac{\sqrt{1-x^2}}{\sqrt{k+1}-x}dx.$ Then $\sum_{k=0}^{99}f(k) =$

$$For convenience let us shift the index $k$ by one, so$\\f(k - 1) = \int^1_{-1} \frac{\sqrt{1 - x^2}}{\sqrt{k} - x} \, dx.$ Here $k \in \mathbb{N}.$

$$Now consider the indefinite integral $I = \int \frac{\sqrt{1 - x^2}}{\sqrt{k} - x} \, dx.$

$\begin{align*}I &= \int \frac{\sqrt{1 - x^2}}{\sqrt{k} - x} \times \frac{\sqrt{1 - x^2}}{\sqrt{1 - x^2}} \, dx\\&= \int \frac{1 - x^2}{(\sqrt{k} - x) \sqrt{ 1 - x^2}} \, dx\\&= \int \frac{(k - x^2) - k + 1}{(\sqrt{k} - x)\sqrt{1 - x^2}} \, dx\\&= \int \frac{k - x^2}{(\sqrt{k} - x)\sqrt{1 - x^2}} \, dx - (k - 1) \int \frac{dx}{(\sqrt{k} - x)\sqrt{1 - x^2}}\\&= J_1 - (k - 1) J_2\end{align*}$

$$For the first integral we have $\\\begin{align*}J_1 &= \int \frac{k - x^2}{(\sqrt{k} - x)\sqrt{1 - x^2}} \, dx\\&= \int \frac{(\sqrt{k} - x)(\sqrt{k} + x)}{(\sqrt{k} - x)\sqrt{1 - x^2}} \, dx\\&= \int \frac{\sqrt{k} + x}{\sqrt{1 - x^2}} \, dx\\&= \sqrt{k} \int \frac{dx}{\sqrt{1 - x^2}} + \int \frac{x}{\sqrt{1 - x^2}} \, dx\\&= \sqrt{k} \sin^{-1} x - \sqrt{1 - x^2} + {\cal{C}}_1\end{align*}$

$$And for the second integral, let $x = \sin \theta, dx = \cos \, d\theta$ so\\\begin{align*}J_2 &= \int \frac{\cos \theta}{(\sqrt{k} - \sin \theta) \sqrt{1 - \sin^2 \theta}} \, d\theta = \int \frac{d\theta}{\sqrt{k} - \sin \theta}\end{align*}$

$Now if a $t$-substitution is used we have $t = \tan \frac{\theta}{2}, d\theta = \frac{2}{1 + t^2} \, dt,$ and $\sin \theta = \frac{2t}{1 + t^2}.$ So one has$

$\begin{align*}J_2 &= 2 \int \frac{dt}{\sqrt{k} (1 + t^2) - 2t}\\&= \frac{2}{\sqrt{k}} \int \frac{dt}{t^2 - \frac{2}{\sqrt{k}} t + 1}\\&= \frac{2}{\sqrt{k}} \int \frac{dt}{\left (t - \frac{1}{\sqrt{k}} \right )^2 + \left (1 - \frac{1}{k} \right )}\\&= \frac{2}{\sqrt{k}} \int \frac{dt}{\left (t - \frac{1}{\sqrt{k}} \right )^2 + \left (\sqrt{\frac{k - 1}{k}} \right )^2}\\&= \frac{2}{\sqrt{k - 1}} \tan^{-1} \left (\frac{\sqrt{k} t - 1}{\sqrt{k - 1}} \right ) + {\cal{C}}_2\\&= \frac{2}{\sqrt{k - 1}} \tan^{-1} \left (\frac{\sqrt{k} \tan \frac{\theta}{2} - 1}{\sqrt{k - 1}} \right ) + {\cal{C}}_2\end{align*}$

$$As $x = \sin \theta$ we have $\cos \theta = \sqrt{1 - x^2}$. Using the result $\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}$ we have $\tan \frac{\theta}{2} = \frac{x}{1 + \sqrt{1 - x^2}}$ and $J_2$ may be written as\\\begin{align*}J_2 &= \frac{2}{\sqrt{k - 1}} \tan^{-1} \left (\frac{\sqrt{k} x - \sqrt{1 - x^2} - 1}{\sqrt{k -1} (\sqrt{1 - x^2} + 1)} \right ) + {\cal{C}}_2.\end{align}$

$$Thus $\\\begin{align*}I = -\sqrt{1 - x^2} + \sqrt{k} \sin^{-1} x - 2 \sqrt{k - 1} \tan^{-1} \left (\frac{\sqrt{k} x - \sqrt{1 - x^2} - 1}{\sqrt{k -1} (\sqrt{1 - x^2} + 1)} \right ) + {\cal{C}}.\end{align}$

$$Now $\\\begin{align*}f(k - 1) &= \int^1_{-1} \frac{\sqrt{1 - x^2}}{\sqrt{k} - x} \, dx\\&= \left [-\sqrt{1 - x^2} + \sqrt{k} \sin^{-1} x - 2 \sqrt{k - 1} \tan^{-1} \left (\frac{\sqrt{k} x - \sqrt{1 - x^2} - 1}{\sqrt{k -1} (\sqrt{1 - x^2} + 1)} \right ) \right ]^1_{-1}\\&= \pi \sqrt{k} - 2 \sqrt{k -1} \left [\tan^{-1} \left (\frac{\sqrt{k} - 1}{\sqrt{k - 1}} \right ) + \tan^{-1} \left (\frac{\sqrt{k} + 1}{\sqrt{k - 1}} \right ) \right ].\end{align*}$

$$If we set $u = \frac{\sqrt{k} - 1}{\sqrt{k - 1}},$ observe that$\\\begin{align*}\frac{1}{u} &= \frac{\sqrt{k - 1}}{\sqrt{k} - 1} \times \frac{\sqrt{k} + 1}{\sqrt{k} + 1} = \frac{\sqrt{k} + 1}{\sqrt{k - 1}},\end{align*}$

$$so that$\\\begin{align*}\tan^{-1} \left (\frac{\sqrt{k} - 1}{\sqrt{k - 1}} \right ) + \tan^{-1} \left (\frac{\sqrt{k} + 1}{\sqrt{k - 1}} \right )&= \tan^{-1}(u) + \tan^{-1} \left (\frac{1}{u} \right ) = \frac{\pi}{2}, \quad \mbox{since} \,\, u > 0, k \geqslant 2.\end{align*}\\$Thus $f(k - 1) = \pi \sqrt{k} - \pi \sqrt{k - 1}, \,\, k \geqslant 2.$

$The case when $k = 1$ needs to be considered separately (this part has been added after the original post). When $k = 1$ we have$

$\begin{align*}f(0) &= \int^1_{-1} \frac{\sqrt{1 - x^2}}{1 - x} \, dx\\&= \int^1_{-1} \frac{\sqrt{1 - x^2}}{1 - x} \times \frac{\sqrt{1 - x^2}}{\sqrt{1 - x^2}}\\&= \int^1_{-1} \frac{1 - x^2}{(1 - x) \sqrt{1 - x^2}} \, dx\\&= \int^1_{-1} \frac{1 + x}{\sqrt{1 - x^2}} \, dx\\&= \int^1_{-1} \frac{dx}{\sqrt{1 - x^2}} + \int^1_{-1} \frac{x}{\sqrt{1 - x^2}} \, dx\\&= \left [\sin^{-1} x \right ]^1_{-1} + \left [-\sqrt{1 - x^2} \right ]^1_{-1}\\&= \pi\\&= \pi (\sqrt{1} - \sqrt{0} )\end{align*}$

$So we have $f(k - 1) = \pi \sqrt{k} - \pi \sqrt{k - 1}, \,\, k \geqslant 1.$\\Finding the required sum is now trivial as the series telescopes. Here$

$\begin{align*}\sum^{99}_{k = 0} f(k)&= \sum^{100}_{k = 1} f(k - 1)\\&= \pi \sum^{100}_{k = 1} (\sqrt{k} - \sqrt{k - 1})\\&= \pi \left [(\sqrt{1} - \sqrt{0}) + (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + \cdots + (\sqrt{99} - \sqrt{98}) + (\sqrt{100} - \sqrt{99}) \right ]\\&= \pi \sqrt{100}\\&= 10 \pi.\end{align*}$

InteGrand · Oct 24, 2015

Re: MX2 2015 Integration Marathon

omegadot said:
$$For convenience let us shift the index $k$ by one, so$\\f(k - 1) = \int^1_{-1} \frac{\sqrt{1 - x^2}}{\sqrt{k} - x} \, dx.$ Here $k \in \mathbb{N}.$

$$Now consider the indefinite integral $I = \int \frac{\sqrt{1 - x^2}}{\sqrt{k} - x} \, dx.$

$\begin{align*}I &= \int \frac{\sqrt{1 - x^2}}{\sqrt{k} - x} \times \frac{\sqrt{1 - x^2}}{\sqrt{1 - x^2}} \, dx\\&= \int \frac{1 - x^2}{(\sqrt{k} - x) \sqrt{ 1 - x^2}} \, dx\\&= \int \frac{(k - x^2) - k + 1}{(\sqrt{k} - x)\sqrt{1 - x^2}} \, dx\\&= \int \frac{k - x^2}{(\sqrt{k} - x)\sqrt{1 - x^2}} \, dx - (k - 1) \int \frac{dx}{(\sqrt{k} - x)\sqrt{1 - x^2}}\\&= J_1 - (k - 1) J_2\end{align*}$

$$For the first integral we have $\\\begin{align*}J_1 &= \int \frac{k - x^2}{(\sqrt{k} - x)\sqrt{1 - x^2}} \, dx\\&= \int \frac{(\sqrt{k} - x)(\sqrt{k} + x)}{(\sqrt{k} - x)\sqrt{1 - x^2}} \, dx\\&= \int \frac{\sqrt{k} + x}{\sqrt{1 - x^2}} \, dx\\&= \sqrt{k} \int \frac{dx}{\sqrt{1 - x^2}} + \int \frac{x}{\sqrt{1 - x^2}} \, dx\\&= \sqrt{k} \sin^{-1} x - \sqrt{1 - x^2} + {\cal{C}}_1\end{align*}$

$$And for the second integral, let $x = \sin \theta, dx = \cos \, d\theta$ so\\\begin{align*}J_2 &= \int \frac{\cos \theta}{(\sqrt{k} - \sin \theta) \sqrt{1 - \sin^2 \theta}} \, d\theta = \int \frac{d\theta}{\sqrt{k} - \sin \theta}\end{align*}$

$Now if a $t$-substitution is used we have $t = \tan \frac{\theta}{2}, d\theta = \frac{2}{1 + t^2} \, dt,$ and $\sin \theta = \frac{2t}{1 + t^2}.$ So one has$

$\begin{align*}J_2 &= 2 \int \frac{dt}{\sqrt{k} (1 + t^2) - 2t}\\&= \frac{2}{\sqrt{k}} \int \frac{dt}{t^2 - \frac{2}{\sqrt{k}} t + 1}\\&= \frac{2}{\sqrt{k}} \int \frac{dt}{\left (t - \frac{1}{\sqrt{k}} \right )^2 + \left (1 - \frac{1}{k} \right )}\\&= \frac{2}{\sqrt{k}} \int \frac{dt}{\left (t - \frac{1}{\sqrt{k}} \right )^2 + \left (\sqrt{\frac{k - 1}{k}} \right )^2}\\&= \frac{2}{\sqrt{k - 1}} \tan^{-1} \left (\frac{\sqrt{k} t - 1}{\sqrt{k - 1}} \right ) + {\cal{C}}_2\\&= \frac{2}{\sqrt{k - 1}} \tan^{-1} \left (\frac{\sqrt{k} \tan \frac{\theta}{2} - 1}{\sqrt{k - 1}} \right ) + {\cal{C}}_2\end{align*}$

$$As $x = \sin \theta$ we have $\cos \theta = \sqrt{1 - x^2}$. Using the result $\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}$ we have $\tan \frac{\theta}{2} = \frac{x}{1 + \sqrt{1 - x^2}}$ and $J_2$ may be written as\\\begin{align*}J_2 &= \frac{2}{\sqrt{k - 1}} \tan^{-1} \left (\frac{\sqrt{k} x - \sqrt{1 - x^2} - 1}{\sqrt{k -1} (\sqrt{1 - x^2} + 1)} \right ) + {\cal{C}}_2.\end{align}$

$$Thus $\\\begin{align*}I = -\sqrt{1 - x^2} + \sqrt{k} \sin^{-1} x - 2 \sqrt{k - 1} \tan^{-1} \left (\frac{\sqrt{k} x - \sqrt{1 - x^2} - 1}{\sqrt{k -1} (\sqrt{1 - x^2} + 1)} \right ) + {\cal{C}}.\end{align}$

$$Now $\\\begin{align*}f(k - 1) &= \int^1_{-1} \frac{\sqrt{1 - x^2}}{\sqrt{k} - x} \, dx\\&= \left [-\sqrt{1 - x^2} + \sqrt{k} \sin^{-1} x - 2 \sqrt{k - 1} \tan^{-1} \left (\frac{\sqrt{k} x - \sqrt{1 - x^2} - 1}{\sqrt{k -1} (\sqrt{1 - x^2} + 1)} \right ) \right ]^1_{-1}\\&= \pi \sqrt{k} - 2 \sqrt{k -1} \left [\tan^{-1} \left (\frac{\sqrt{k} - 1}{\sqrt{k - 1}} \right ) + \tan^{-1} \left (\frac{\sqrt{k} + 1}{\sqrt{k - 1}} \right ) \right ].\end{align*}$

$$If we set $u = \frac{\sqrt{k} - 1}{\sqrt{k - 1}},$ observe that$\\\begin{align*}\frac{1}{u} &= \frac{\sqrt{k - 1}}{\sqrt{k} - 1} \times \frac{\sqrt{k} + 1}{\sqrt{k} + 1} = \frac{\sqrt{k} + 1}{\sqrt{k - 1}},\end{align*}$

$$so that$\\\begin{align*}\tan^{-1} \left (\frac{\sqrt{k} - 1}{\sqrt{k - 1}} \right ) + \tan^{-1} \left (\frac{\sqrt{k} + 1}{\sqrt{k - 1}} \right )&= \tan^{-1}(u) + \tan^{-1} \left (\frac{1}{u} \right ) = \frac{\pi}{2}, \quad \mbox{since} \,\, u > 0.\end{align*}$

$Thus$\\f(k - 1) = \pi \sqrt{k} - \pi \sqrt{k - 1}.\\$Finding the required sum is now trivial as the series telescopes. Here$

$\begin{align*}\sum^{99}_{k = 0} f(k)&= \sum^{100}_{k = 1} f(k - 1)\\&= \pi \sum^{100}_{k = 1} (\sqrt{k} - \sqrt{k - 1})\\&= \pi \left [(\sqrt{1} - \sqrt{0}) + (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + \cdots + (\sqrt{99} - \sqrt{98}) + (\sqrt{100} - \sqrt{99}) \right ]\\&= \pi \sqrt{100}\\&= 10 \pi.\end{align*}$

Nice solution. I think it would have saved some working though to just do the integrals as definite ones from the start and change the limits as you go for the substitutions.

lita1000 · Oct 24, 2015

Re: MX2 2015 Integration Marathon

omegadot said:
$$For convenience let us shift the index $k$ by one, so$\\f(k - 1) = \int^1_{-1} \frac{\sqrt{1 - x^2}}{\sqrt{k} - x} \, dx.$ Here $k \in \mathbb{N}.$

$$Now consider the indefinite integral $I = \int \frac{\sqrt{1 - x^2}}{\sqrt{k} - x} \, dx.$

$\begin{align*}I &= \int \frac{\sqrt{1 - x^2}}{\sqrt{k} - x} \times \frac{\sqrt{1 - x^2}}{\sqrt{1 - x^2}} \, dx\\&= \int \frac{1 - x^2}{(\sqrt{k} - x) \sqrt{ 1 - x^2}} \, dx\\&= \int \frac{(k - x^2) - k + 1}{(\sqrt{k} - x)\sqrt{1 - x^2}} \, dx\\&= \int \frac{k - x^2}{(\sqrt{k} - x)\sqrt{1 - x^2}} \, dx - (k - 1) \int \frac{dx}{(\sqrt{k} - x)\sqrt{1 - x^2}}\\&= J_1 - (k - 1) J_2\end{align*}$

$$For the first integral we have $\\\begin{align*}J_1 &= \int \frac{k - x^2}{(\sqrt{k} - x)\sqrt{1 - x^2}} \, dx\\&= \int \frac{(\sqrt{k} - x)(\sqrt{k} + x)}{(\sqrt{k} - x)\sqrt{1 - x^2}} \, dx\\&= \int \frac{\sqrt{k} + x}{\sqrt{1 - x^2}} \, dx\\&= \sqrt{k} \int \frac{dx}{\sqrt{1 - x^2}} + \int \frac{x}{\sqrt{1 - x^2}} \, dx\\&= \sqrt{k} \sin^{-1} x - \sqrt{1 - x^2} + {\cal{C}}_1\end{align*}$

$$And for the second integral, let $x = \sin \theta, dx = \cos \, d\theta$ so\\\begin{align*}J_2 &= \int \frac{\cos \theta}{(\sqrt{k} - \sin \theta) \sqrt{1 - \sin^2 \theta}} \, d\theta = \int \frac{d\theta}{\sqrt{k} - \sin \theta}\end{align*}$

$Now if a $t$-substitution is used we have $t = \tan \frac{\theta}{2}, d\theta = \frac{2}{1 + t^2} \, dt,$ and $\sin \theta = \frac{2t}{1 + t^2}.$ So one has$

$\begin{align*}J_2 &= 2 \int \frac{dt}{\sqrt{k} (1 + t^2) - 2t}\\&= \frac{2}{\sqrt{k}} \int \frac{dt}{t^2 - \frac{2}{\sqrt{k}} t + 1}\\&= \frac{2}{\sqrt{k}} \int \frac{dt}{\left (t - \frac{1}{\sqrt{k}} \right )^2 + \left (1 - \frac{1}{k} \right )}\\&= \frac{2}{\sqrt{k}} \int \frac{dt}{\left (t - \frac{1}{\sqrt{k}} \right )^2 + \left (\sqrt{\frac{k - 1}{k}} \right )^2}\\&= \frac{2}{\sqrt{k - 1}} \tan^{-1} \left (\frac{\sqrt{k} t - 1}{\sqrt{k - 1}} \right ) + {\cal{C}}_2\\&= \frac{2}{\sqrt{k - 1}} \tan^{-1} \left (\frac{\sqrt{k} \tan \frac{\theta}{2} - 1}{\sqrt{k - 1}} \right ) + {\cal{C}}_2\end{align*}$

$$As $x = \sin \theta$ we have $\cos \theta = \sqrt{1 - x^2}$. Using the result $\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}$ we have $\tan \frac{\theta}{2} = \frac{x}{1 + \sqrt{1 - x^2}}$ and $J_2$ may be written as\\\begin{align*}J_2 &= \frac{2}{\sqrt{k - 1}} \tan^{-1} \left (\frac{\sqrt{k} x - \sqrt{1 - x^2} - 1}{\sqrt{k -1} (\sqrt{1 - x^2} + 1)} \right ) + {\cal{C}}_2.\end{align}$

$$Thus $\\\begin{align*}I = -\sqrt{1 - x^2} + \sqrt{k} \sin^{-1} x - 2 \sqrt{k - 1} \tan^{-1} \left (\frac{\sqrt{k} x - \sqrt{1 - x^2} - 1}{\sqrt{k -1} (\sqrt{1 - x^2} + 1)} \right ) + {\cal{C}}.\end{align}$

$$Now $\\\begin{align*}f(k - 1) &= \int^1_{-1} \frac{\sqrt{1 - x^2}}{\sqrt{k} - x} \, dx\\&= \left [-\sqrt{1 - x^2} + \sqrt{k} \sin^{-1} x - 2 \sqrt{k - 1} \tan^{-1} \left (\frac{\sqrt{k} x - \sqrt{1 - x^2} - 1}{\sqrt{k -1} (\sqrt{1 - x^2} + 1)} \right ) \right ]^1_{-1}\\&= \pi \sqrt{k} - 2 \sqrt{k -1} \left [\tan^{-1} \left (\frac{\sqrt{k} - 1}{\sqrt{k - 1}} \right ) + \tan^{-1} \left (\frac{\sqrt{k} + 1}{\sqrt{k - 1}} \right ) \right ].\end{align*}$

$$If we set $u = \frac{\sqrt{k} - 1}{\sqrt{k - 1}},$ observe that$\\\begin{align*}\frac{1}{u} &= \frac{\sqrt{k - 1}}{\sqrt{k} - 1} \times \frac{\sqrt{k} + 1}{\sqrt{k} + 1} = \frac{\sqrt{k} + 1}{\sqrt{k - 1}},\end{align*}$

$$so that$\\\begin{align*}\tan^{-1} \left (\frac{\sqrt{k} - 1}{\sqrt{k - 1}} \right ) + \tan^{-1} \left (\frac{\sqrt{k} + 1}{\sqrt{k - 1}} \right )&= \tan^{-1}(u) + \tan^{-1} \left (\frac{1}{u} \right ) = \frac{\pi}{2}, \quad \mbox{since} \,\, u > 0.\end{align*}$

$Thus$\\f(k - 1) = \pi \sqrt{k} - \pi \sqrt{k - 1}.\\$Finding the required sum is now trivial as the series telescopes. Here$

$\begin{align*}\sum^{99}_{k = 0} f(k)&= \sum^{100}_{k = 1} f(k - 1)\\&= \pi \sum^{100}_{k = 1} (\sqrt{k} - \sqrt{k - 1})\\&= \pi \left [(\sqrt{1} - \sqrt{0}) + (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + \cdots + (\sqrt{99} - \sqrt{98}) + (\sqrt{100} - \sqrt{99}) \right ]\\&= \pi \sqrt{100}\\&= 10 \pi.\end{align*}$

When did you sit your HSC?

omegadot · Oct 24, 2015

Re: MX2 2015 Integration Marathon

lita1000 said:
When did you sit your HSC?

Many moons ago - 1990

omegadot · Oct 24, 2015

Re: MX2 2015 Integration Marathon

omegadot said:
$\int \left (\frac{x}{x\sin x + \cos x} \right )^2\mbox{d}x$

$$In addition to juantheron's solution, I will show how to evaluate this integral using three other methods. Let$\\I = \int \frac{x^2}{(x \sin x + \cos x)^2} \, dx.$

$\textbf{Method I}$

$$Observe that $\frac{d}{dx} (x \sin x + \cos x) = x \cos x.$ So$

$\begin{align*}I &= \int \frac{x \cos x}{(x \sin x + \cos x)^2} \cdot \frac{x}{\cos x} \, dx\\&= -\frac{1}{x \sin x + \cos x} \cdot \frac{x}{\cos x} + \int \frac{1}{x \sin x + \cos x} \cdot \frac{\cos x + x \sin x}{\cos x} \, dx \quad \mbox{(by parts)}\\&= -\frac{x}{\cos x (x \sin x + \cos x)} + \int \frac{dx}{\cos^2 x}\\&= -\frac{x}{\cos x(x \sin x + \cos x)} + \int \sec^2 x \, dx\\&= -\frac{x}{\cos x(x \sin x + \cos x)} +\tan x + \cal{C}\\&=\frac{-x + \sin x (x \sin x + \cos x)}{\cos x (x \sin x + \cos x)} + \cal{C}\\&=\frac{-x \cos^2 x + \sin x \cos x}{\cos x(x \sin x + \cos x)} + \cal{C}\\&= \frac{\sin x - x \cos x}{x \sin x + \cos x} + \cal{C}\end{align*}$

$\textbf{Method II}$

$$As the integrand consists of a fraction with a squared term in the denominator, if the indefinite integral is to have an elementary solution one may think of the quotient rule, namely $\frac{d}{dx} \left (\frac{u}{v} \right ) = \frac{u' v - v' u}{v^2}$ where $u$ and $v$ are functions of $x.$

$So for the integrand we have$\\\begin{align*}\frac{x^2}{(x \sin x + \cos x)^2} &= \frac{u' (x \sin x + \cos x) - (x \sin x + \cos x)' u}{(x \sin x + \cos x)^2} = \frac{u' \cdot (x \sin x + \cos x) - x \cos x \cdot u}{(x \sin x + \cos x)^2}\end{align*}\\$For the numerator, it can be rewritten as follows$

$\begin{align*}x^2 &= x^2 \sin^2 x + x^2 \cos^2 x\\&=x^2 \sin^2 x + x \sin x \cos x - x \sin x \cos x + x^2 \cos^2 x\\&= x \sin x (x \sin x + \cos x) - x \cos x (\sin x - x \cos x),\end{align*}\\$so we se that $u = \sin x - x \cos x.$ Thus$\\\frac{x^2}{(x \sin x + \cos x)^2} = \frac{d}{dx} \left (\frac{\sin x - x \cos x}{x \sin x + \cos x} \right ),$ and the integral becomes$$

$\begin{align*}I &= \int \frac{d}{dx} \left (\frac{\sin x - x \cos x}{x \sin x + \cos x} \right ) \, dx = \frac{\sin x - x \cos x}{x \sin x + \cos x} + \cal{C}\end{align*}\\$as expected.$$

$\textbf{Method III}$

$Let $x = \tan u,$ so $dx = \sec^2 u \, du.$ Thus$

$\begin{align*}I &= \int \frac{\tan^2 u \sec^2 u}{\left [\tan u \sin (\tan u) + \cos (\tan u) \right ]^2} \, du\\&= \int \frac{\tan^2 u \sec^2 u \cos^2 u}{\left [\sin u \sin (\tan u) + \cos (\tan u) \cos u \right ]^2} \, du\\&= \int \frac{\tan^2 u}{\cos^2 (\tan u - u)} \, du\end{align*}$

$And let $t = \tan u - u, dt = (\sec^2 u - 1) \, du = \tan^2 u \, du.$ So we have$

$\begin{align*}I &= \int \frac{dt}{\cos^2 t}\\&= \int \sec^2 t \, dt\\&= \tan t + \cal{C}\\&= \tan(\tan u - u) + \cal{C}\\&= \tan(x - \tan^{-1} x) + \cal{C}\\&= \frac{\tan x - x}{1 + x \tan x} + \cal{C}\\&= \frac{\sin x - x \cos x}{\cos x + x \sin x} + \cal{C},\end{align*}\\$as expected.$$

kawaiipotato · Oct 24, 2015

Re: MX2 2015 Integration Marathon

juantheron said:
$(2)\; Let $I_{n} = \int_{0}^{\frac{\pi}{2}}x(\cos x+\sin x)^ndx\;,$ Then value of $\frac{101I_{101}-\frac{\pi}{2}}{I_{99}} =$

$I_{n} = \int_{0}^{\frac{\pi}{2}}x(\cos x+\sin x)^ndx$

$I_{n} = \int_{0}^{\frac{\pi}{2}}(\frac{\pi}{2} - x)(\cos x+\sin x)^ndx$

$2I_{n} = \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} dx$

$I_{n} = \frac{\pi}{4} \left[x \right]_{0}^{\frac{\pi}{2}}$

$=\frac{\pi ^2 }{8}$

$\therefore \frac{101I_{101}-\frac{\pi}{2}}{I_{99}} = \frac{101 * \frac{\pi ^2 }{8} - \frac{\pi}{2}}{\frac{\pi ^{2}}{8}}$

$= \frac{101 \pi - 4}{ \pi}$

Weird, WolframAlpha is giving me different values for the integral when I change the values of n. It should just be pi^2 / 8 right?
lol nvm added it wrong

InteGrand · Oct 24, 2015

Re: MX2 2015 Integration Marathon

kawaiipotato said:
$I_{n} = \int_{0}^{\frac{\pi}{2}}(\frac{\pi}{2} - x)(\cos x+\sin x)^ndx$

$2I_{n} = \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} dx$

I think you made a mistake expanding when going from the top line here to the next.

InteGrand · Oct 24, 2015

Re: MX2 2015 Integration Marathon

omegadot said:
$$(b)$ \int \left (\sin^{-1} x \right )^3 dx.$

$Let $I_n =\int \left(\sin^{-1} x\right)^n \text{ d}x$, for $n\in \mathbb{N}$. We want to find $I_3$. Note that$

$\begin{align*}I_1 &= \int \sin^{-1}x \text{ d}x \\ &= x\sin^{-1} x - \int \frac{x}{\sqrt{1-x^2}}\text{ d}x\\ &= x \sin^{-1}x + \sqrt{1-x^2}+\mathcal{C}. \end{align*}$

$We develop a recurrence for $I_n$. Let $\theta = \sin^{-1}x$, so $x = \sin \theta$, and $\theta \in \left[-\frac{\pi}{2},\frac{\pi}{2} \right]$, so $\cos \theta \geq 0 \Rightarrow \cos \theta = \sqrt{1-\sin^2 \theta}=\sqrt{1-x^2}$. Also, $\mathrm{d}x = \cos \theta \text{ d}\theta$, so we have using integration by parts twice$

$\begin{align*} I_n &= \int \theta ^n \cos \theta \text{ d}\theta \\ &= \sin \theta \cdot \theta ^n - \int \sin \theta \times n \theta^{n-1}\text{ d}\theta \\ &= \theta^n \sin \theta -n \int \theta ^{n-1}\sin \theta \text{ d}\theta \\ &= \theta^n \sin \theta -n \left[-\cos \theta \cdot \theta ^{n-1}+\int \cos \theta \cdot (n-1) \theta ^{n-2}\text{ d} \theta \right]\\ &= \theta ^n \sin \theta + n\theta ^{n-1}\cos \theta - n(n-1)\int \theta ^{n-2}\cos \theta \text{ d}\theta \\&= \theta ^n \sin \theta + n\theta ^{n-1}\cos \theta - n(n-1)I_{n-2}\quad (n\geq 2). \end{align*}$

$Therefore, we have$

$\begin{align*} I_3 &= \theta ^3 \sin \theta + 3 \theta ^2 \cos \theta - 6 I_1 \\ &= \left( \sin ^{-1}x \right)^3 \cdot x + 3\left( \sin ^{-1}x \right)^2 \cdot \sqrt{1-x^2} - 6 \left(x \sin ^{-1}x + \sqrt{1-x^2} \right)+ \mathcal{K} \\ &= x\sin^{-1}x \left[\left(\sin^{-1}x \right)^2 - 6 \right] + 3 \sqrt{1-x^2}\left[\left(\sin^{-1}x\right)^2 -2 \right]+ \mathcal{K}.\end{align*}$

kawaiipotato · Oct 24, 2015

Re: MX2 2015 Integration Marathon

$I_{n} = \int_{0}^{\frac{\pi}{2}}(\frac{\pi}{2} - x)(\cos x+\sin x)^ndx$

$2I_{n} = \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} dx$

$I_{n} = \frac{\pi}{4} \sqrt{2^n} \int_{0}^{\frac{\pi}{2}} (sin(x + \frac{\pi}{4})^n dx$

$$Consider$ \int_{0}^{\frac{\pi}{2}} (\sin (x + \frac{\pi}{4})^n = J_{n}$

$$u$ = (\sin (x + \frac{\pi}{4})^{n-1}$ ... $dv$ = \sin (x + \frac{\pi}{4})$dx$

$du$ = (n-1)cos(x + \frac{\pi}{4})(\sin (x + \frac{\pi}{4})^{n-2} $dx$ ... $v = -\cos (x + \frac{\pi}{4})$

$\therefore J_{n} = -\left[\cos (x + \frac{\pi}{4})(\sin (x + \frac{\pi}{4})^{n-1}\right]_{0}^{\frac{\pi}{2}} + (n-1) \int_{0}^{\frac{\pi}{2}} cos^{2}(x + \frac{\pi}{4})(\sin (x + \frac{\pi}{4})^{n-2} dx$

$J_{n} = 2 \left(\frac{1}{\sqrt 2}\right)^{n+1} + (n-1)J_{n-2} - (n-1)J_{n}$

$J_{n} = \frac{2 \left(\frac{1}{\sqrt 2}\right)^{n+1} + (n-1)J_{n-2}}{n}$

$\therefore I_{n} = \frac{\pi}{4}* \sqrt{2^n} * \frac{2 \left(\frac{1}{\sqrt 2}\right)^{n+1} + (n-1)J_{n-2}}{n}$

omegadot · Oct 24, 2015

Re: MX2 2015 Integration Marathon

$$(a) $ \int \left (\frac{\tan^{-1} x}{x - \tan^{-1} x} \right )^2 \,dx \qquad $(b) $ \int^1_{-1} \tan^{-1} \left ({\rm{e}}^{\sin x} \right ) \, dx.$

InteGrand · Oct 24, 2015

Re: MX2 2015 Integration Marathon

omegadot said:
$$(a) $ \int \left (\frac{\tan^{-1} x}{x - \tan^{-1} x} \right )^2 \,dx \qquad $(b) $ \int^1_{-1} \tan^{-1} \left ({\rm{e}}^{\sin x} \right ) \, dx.$

$For (b), let the integral be $I$. Splitting the integral yields $I=\int _{-1} ^0 \tan^{-1} \left(\mathrm{e}^{\sin x} \right)\text{ d}x + \int_0 ^1 \tan^{-1} \left(\mathrm{e}^{\sin x}\right)\text{ d}x$. For the first integral, let $u=-x$, so $\sin x = \sin (-u) = -\sin u$, $\mathrm{d}u = -\text{d}x$, and$

$\begin{align*} I &= \int _{1} ^0 \tan^{-1} \left(\mathrm{e}^{-\sin u} \right) \cdot (-1) \text{ d}u + \int_0 ^1 \tan^{-1} \left( \mathrm{e}^{\sin x} \right)\text{ d}x \\ &= \int _0 ^1 \tan ^{-1} \left( \frac{1}{\mathrm{e}^{\sin u}} \right) \text{ d}u +\int_0 ^1 \tan^{-1} \left(\mathrm{e}^{\sin x}\right) \text{ d}x \\ &= \int _0 ^1 \left(\tan^{-1} \left( \frac{1}{\mathrm{e}^{\sin t}} \right) + \tan^{-1} \left( \mathrm{e}^{\sin t} \right) \right) \text{ d}t \\ &= \int _0 ^1 \frac{\pi}{2}\text{ d}t \quad (\text{as the arguments of the inverse tangents are positive})\\ &= \frac{\pi}{2}. \end{align*}$

juantheron · Oct 24, 2015

Re: MX2 2015 Integration Marathon

$Let $I = \int\left(\frac{\tan^{-1} x}{x-\tan^{-1} x}\right)^2 dx =\int\left(\frac{\tan^{-1} x}{\tan^{-1} x-x}\right)^2 dx$

$So we can write $I = \int\left(1+\frac{x}{\tan^{-1} x-x}\right)^2dx$

$So we get $I =\int 1dx+\int\frac{2x}{\tan^{-1} x-x}dx+\int\left(\frac{x}{\tan^{-1}x-x}\right)^2dx$

$Now $\left(\frac{1}{\tan^{-1}x-x}\right)' = -\frac{\frac{1}{1+x^2}-1}{\left(\tan^{-1}x-x\right)^2} = \frac{1}{1+x^2}\left(\frac{x}{\tan^{-1}x-x}\right)^2$

$So $\left(\frac{x}{\tan^{-1}x-x}\right)^2 = (1+x^2)\cdot \left(\frac{1}{\tan^{-1}x-x}\right)'$

$So $\int \left(\frac{x}{\tan^{-1}x-x}\right)^2dx = \int (1+x^2)\cdot \left(\frac{1}{\tan^{-1}x-x}\right)'$

$So we get $ = (1+x^2)\cdot \frac{1}{\tan^{-1}x-x}-\int\frac{2x}{\tan^{-1}x-x}dx$

$Now $I = \int 1dx+\int\frac{2x}{\tan^{-1}x-x}dx+(1+x^2)\cdot \frac{1}{\tan^{-1}x-x}-\int\frac{2x}{\tan^{-1}x-x}dx$

$So Integral $I = \int\left(\frac{\tan^{-1} x}{x-\tan^{-1} x}\right)^2 dx = x+\frac{1+x^2}{\tan^{-1}x-x}+\mathcal{C}$

juantheron · Oct 24, 2015

Re: MX2 2015 Integration Marathon

$Let, $\text{V(n)}=\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin^2 x} \text{d}x$

$and $\text{J}= \text{V(n) - V(n-1)}=\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\sin^2nx-\sin^2(n-1)x}{\sin^2 x} \text{d}x$

$=\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\sin(2n-1)x\times\sin x}{\sin^2 x} \text{d}x$

$$**[Using $(\sin^2a - \sin^2b) = \sin(a+b)\times \sin(a-b)$]**$

$=\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\sin(2n-1)x}{\sin x} \text{d}x$

$Now, consider $\text{S}=\cos (2x) + \cos(4x) + \cos(6x) +......+ \cos(2n-2)x$

$So we get $S = \frac {\sin(n-1)x \times \cos nx}{\sin x}$

$=\dfrac {2\sin(n-1)x \times \cos nx}{2\sin x}$

$=\dfrac {\sin(2n-1)x - \sin x}{2\sin x}$

$\implies \text{J}=\displaystyle \int_{0}^{\frac{\pi}{2}}(2\text{S}+1) \: \text{d}x=\dfrac{\pi}{2}$

$\implies \text{V(1), V(2), V(3) ..... V(n)}$ form an **Arithmetic Progression.**$

$Since $\text{V(1)}=\dfrac{\pi}{2}$ , $\text{V(n)}=\boxed{\dfrac{n\pi}{2}}$

$Now $\sum^{100}_{r=1}V_{r} = \frac{\pi}{2}\sum^{100}_{r=1}r = \frac{\pi}{2}\times \frac{100 \times 101}{2} = 25\times 101 \times \pi$

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