Solving a Tricky Integral: 0 to $\pi/4$

[G01]

\int_0^{\pi/4} \frac{x\sin(x)}{cos^3(x)}dx

I thought I could do this integral by parts but I keep getting it wrong and I can't find my mistake.

\int_0^{\pi/4} x\tan(x)\sec^2(x) dx

let u = xtanx , dv = sec^2x dx

du = xsec^2x+tanx , v = tanx

x\tan^2(x)|_0^{\pi/4} - \int_0^{\pi/4} x\tan(x)\sec^2(x) dx - \int_0^{\pi/4} \tan^2(x) dx

so:

\int_0^{\pi/4} x\tan(x)\sec^2(x) dx = \frac{x\tan^2(x)|_0^{\pi/4} - \int_0^{\pi/4} \tan^2(x) dx}{2}

x\tan^2(x)|_0^{\pi/4} - x|_0^{\pi/4} + \tan(x)|_0^{\pi/4}

This evaluates out to 1 but that's not the answer. Thank you for your help.

 

[0rthodontist]

Let u = x, dv = tan x sec^2 x. You can integrate the second by substitution, and the x will drop out by differentiation.

Your error is in the last step, where you have -x + tan x. It should be tan x - x. Also you forgot to divide by 2.

 

[G01]

thanks alot