Quick Integration Question (1 Viewer)

[bcd]

Hey, can anybody please help me with an Integration question. Its from the Cambridge book, diagnostic test, p175

Here it is:

a) Show that [integral of f(X) with limits a and 0] = [int of f(a-x) with limits a to 0]
This bit is fine.

b) Hence show that:
{ Int of [ (cosX) / (cosX +sinX) ] with limits pi/2 and 0 } = pi/4

Thanks
BCD

 

[香港!]

Hey, can anybody please help me with an Integration question. Its from the Cambridge book, diagnostic test, p175

Here it is:

a) Show that [integral of f(X) with limits a and 0] = [int of f(a-x) with limits a to 0]
This bit is fine.

b) Hence show that:
{ Int of [ (cosX) / (cosX +sinX) ] with limits pi/2 and 0 } = pi/4

Thanks
BCD

I=int. cosx\(cosx+sinx) dx {from 0 to pi\2}
=int. cos(pi\2-x) \ (cos(pi\2-x)+sin(pi\2-x) ) dx
=int. sinx\(sinx+cosx) dx
int. sinx\(sinx+cosx) dx+int. cosx\(cosx+sinx) dx=2I
2I=int. (sinx+cosx)\(sinx+cosx) dx=int. 1 dx=[x ] (from 0 to pi\2)
=pi\2
.: I=pi\4

 

[bcd]

Thank you!

 

[Stefano]

I=int. cosx\(cosx+sinx) dx {from 0 to pi\2}
=int. cos(pi\2-x) \ (cos(pi\2-x)+sin(pi\2-x) ) dx
=int. sinx\(sinx+cosx) dx
int. sinx\(sinx+cosx) dx+int. cosx\(cosx+sinx) dx=2I
2I=int. (sinx+cosx)\(sinx+cosx) dx=int. 1 dx=[x ] (from 0 to pi\2)
=pi\2
.: I=pi\4

The integration master strikes again...! Who will be next I wonder?

 

[香港!]

LoL dude I aint no integration master... maybe she just hasn't bothered to open up her text book... this question is on almost every 4U text book that has integration in it

 

[Stefano]

Lol, I feel stupid now because I just attempted this and couldn't get it. I feel even worse for not being able to follow your solution either Mr.HK!

-Why do u change the arguments to (pi/2-x) ??
-How do u get from line2 to line3 ??

-is there another way to do this ?

 

[香港!]

Lol, I feel stupid now because I just attempted this and couldn't get it. I feel even worse for not being able to follow your solution either Mr.HK!

-Why do u change the arguments to (pi/2-x) ??
-How do u get from line2 to line3 ??

-is there another way to do this ?

I used the property:
int. f(x) dx {from 0 to a}=int. f(a-x) dx {from 0 to a}
bcd proved this in part one of the question...

"-Why do u change the arguments to (pi/2-x) ??"

the limits were from 0 to pi\2
I just replaced every x with (a-x), i.e. (pi\2-x)

"-How do u get from line2 to line3 ??"
cos (pi\2-x)=sinx
sin(pi\2-x)=cosx
Is that what you're looking for?

 

[Dumsum]

You change it to pi/2 - x because that's the identity you proved in part i)

Line 2 to line 3 is just trig identities
cos(pi/2 - x) = sinx
sin(pi/2 - x) = cosx

Edit: too slow

 

[Stefano]

Thanks. I get it. (Except the trig identities Dumdum mentioned...)

I don't like it though. I was hoping you could use a nice little substitution or something.

 

[Antwan23q]

do u have the arnoalds 4 unit text book? at the end of the chapter of integration, there is a section on integrations like this

 

[Stefano]

do u have the arnoalds 4 unit text book? at the end of the chapter of integration, there is a section on integrations like this

No, I do not have that book and have never seen integrals like this before. I'm just hoping it doesn't come up.

 

[Slidey]

Patel also covers it. It's a pretty standard question, but I wouldn't call it common.

It's actually rather easy to master, so have a go at some more questions like it anyway.