That wouldn't happen to be the MATH1241 Calculus book..?
who needs a substitution?Yep!
Now:
Can be done in about five easy lines with a smart substitution.
Did you recently acquire a tablet? If not, when is your birthday?
Has anyone solved this question yet?Find a recursion relation for the following sequence of integrals:
where n is a non-negative integer.
Hence evaluate.
I did this question a couple days ago and got I(n)=[e^x*sin^n(x)-ne^x*sin^(n-1)(x)cos(x)+n(n-1)I(n-2)]/[n^2+1] for the first part.
Well I haven't even attempted that part but it doesn't look too difficult. I thought you were talking about UMADBRO's integral.
Now, note that:
therefore our integral can be written as
therefore
EDIT: LOL FORGOT TO INTEGRATE.
could someone explain to me what happened between the 1st line and the 2nd line?U MAD BRO keep your integrals at the level of the MX2 integration topic. If you are interested in how to do these yourself, do outside research / ask in extracurricular. This thread is an inappropriate place to post them.
@Rolpsy nice question!
\,dx=\int_0^{\pi/2}(\pi/2-x)\tan(x)\,dx\\ \Rightarrow 2I=\int_0^{\pi/2}\frac{2x\cos^2(x)+2(\pi/2-x)\sin^2(x)}{2\sin(x)\cos(x)}\, dx\\=\int_0^{\pi/2}\frac{2x\cos(2x)+\frac{\pi}{2}(1-\cos(2x))}{\sin(2x)}\,dx\\\Rightarrow 4I=\int_0^{\pi}\frac{x\cos(x)+\frac{\pi}{2}(1-\cos(x))}{\sin(x)}\, dx\\=\int_0^{\pi/2}x\cot(x)\,dx+\frac{\pi}{2}\int_0^{\pi/2}\csc(x)-\cot(x)\,dx+\int_{\pi/2}^\pi (x-\pi)\cot(x)\,dx+\frac{\pi}{2}\int_{\pi/2}^\pi\csc(x)+\cot(x)\,dx\\=2I+\pi\int_0^{\pi/2}\csc(x)-\cot(x)\,dx\\=2I-\pi\log|1+\cos(x)|]^{\pi/2}_0\\\Rightarrow I=\frac{\pi}{2}\log(2).)
He added the definite integrals of
