I'm very new for calculus.
please solve this problem. thank you very much.

2. Originally Posted by ronaldkaka

I'm very new for calculus.
please solve this problem. thank you very much.

Multiply top and bottom by ( sin^7 x - cos ^7 x)

you get on the bottom sin^14 x - cos^14 x

cos^2 x = 1 -sin^2x

Use this identity then finish of the question.

3. Thank you very much Mr.╔(σ_σ)╝.

4. Originally Posted by ronaldkaka
Thank you very much Mr.╔(σ_σ)╝.
I gave you a useless step, sorry i wasn't thinking about it.

Anyway i'm going to suggest something radical and an easy way out.

But i'm not sure if you are allowed to leave your answers in terms of non elementary fuctions.

Divide everything by cos^7 x

you get $\frac{tan^{7} x}{ 1 + tan^{7} x} = 1 - \frac{1}{1 + tan^{7}x}$

use a substitution u = tanx

du = sec^2 x = ( u^2 - 1)

you get

$\frac{1}{ (u^{2} -1)( 1+u^{7})}$ you can do this

$\frac{1}{(u-1)(u+1)(1+u^{7})}$

Partial fractions would be appropriate.

If you get something like

$\frac{Au+b}{(1+u^{7})}$

You may have to use the maclaurin series or do some more algrabra. I didn't try it so i'm not sure what you would end up with.

Good luck though. The algebra seems disgusting.

I popped it into an online integrator and it doesn't look pretty.
Wolfram Mathematica Online Integrator

5. Great.
Thank you very much ╔(σ_σ)╝.

6. I think you might get stuck with the partial fraction.

$\frac{1}{(u^{2}-1)(u^{7} +1)}$ =

$\frac{-3x^{4}+2x^{3} -4x^{2} +2x -3}{7( x^{6}-x^{5}+x^{4} -x^{3} +x^{2}-x +1)} + \frac{1}{4(x-1)}- \frac{1}{4(x+1)} - \frac{1}{14(x+1)^{2}}$

Did it with pc.

it's a very standard integral: do the substitution $x \to \frac{\pi}{2} - x$ and then add the two integrals together to get $\int_0^{\frac{\pi}{2}} dx=\frac{\pi}{2}.$ so the value of the integral is $\frac{\pi}{4}.$