Question

question
http://gyazo.com/1ad26b5998882098c66d7b4115b73ad6

Answer 1

is this right?
http://gyazo.com/8e89a134072f555cd71f313ad350f788

Answer 2

\[
\int_a^b f(g(u))\;du\neq F(g(b))-F(g(a))
\]

Answer 3

However:\[
\int_a^b f(g(u))g'(u)\;du=\int_a^b f(g)\;dg=F(b)-F(a)
\]

Answer 4

oh woops, so i let u = cosx for the first integral?

Answer 5

Well, we start out with: \[
\int_0^{\pi/2}f(\cos(x))\;dx=\int_0^{\pi/2}f(\sin(x))\;dx
\]right?

Answer 6

yes

Answer 7

prove the first integral first, then the next?

Answer 8

or there another way of proving it?

Answer 9

No, I think you're suppose to use the equation to evaluate it.

Answer 10

Consider that: \[
\sin^2(x)+\cos^2(x)=1
\]

Answer 11

it says if f is continuous prove that ....

Answer 12

\[
\forall x\in (0,\pi/2)\quad \cos(x) = \sqrt{1-\sin^2(x)}
\]

Answer 13

Wait you have to prove it is true?!

Answer 14

from the question, if i didnt read it wrongly..
http://gyazo.com/1ad26b5998882098c66d7b4115b73ad6

Answer 15

I guess you need to use \[
\cos(x) = \sin(\pi/2-x)
\]

Answer 16

Maybe \(u=\pi/2-x\) would work here.

Answer 17

That means \(du=-dx\) and \(x=\pi/2-u\) so:\[
\int_0^{\pi/2}f(\cos x)\;dx = \int_{\pi/2}^0f(\cos(\pi/2-u))\;(-du)
\]

Answer 18

@mathsnerd101 Does that help?

Answer 19

im trying it now

Answer 20

No, you never actually integrate it dude. Come on.

Answer 21

\(\color{blue}{\text{Originally Posted by}}\) @wio
I guess you need to use \[
\cos(x) = \sin(\pi/2-x)
\]
\(\color{blue}{\text{End of Quote}}\)
\(\color{blue}{\text{Originally Posted by}}\) @wio
Maybe \(u=\pi/2-x\) would work here.
\(\color{blue}{\text{End of Quote}}\)
\(\color{blue}{\text{Originally Posted by}}\) @wio
That means \(du=-dx\) and \(x=\pi/2-u\) so:\[
\int_0^{\pi/2}f(\cos x)\;dx = \int_{\pi/2}^0f(\cos(\pi/2-u))\;(-du)
\]
\(\color{blue}{\text{End of Quote}}\)

Answer 22

Eventually you'll get to \[
=\int_0^{\pi/2}f(\sin x)\;dx
\]

Answer 23

by simplifying?

Answer 24

Yes.

Answer 25

Use trig identities.

Answer 26

\[
\int_a^b=-\int _b^a
\]

Answer 27

Are you still lost?

Answer 28

ye

Answer 29

\[\begin{split}
\int_0^{\pi/2}f(\cos x)\;dx
&= \int_{\pi/2}^0f(\cos(\pi/2-u))\;(-du) \\
&= -\int_{\pi/2}^0f(\cos(\pi/2-u))\;du \\
&= \int_0^{\pi/2}f(\cos(\pi/2-u))\;du \\
&= \int_0^{\pi/2}f(\sin u)\;du \\
\end{split}\]

Answer 30

then do i change the \(u\) back to \(x\)?

Answer 31

Yes.

Answer 32

ohh, i got it, thanks!!