# Trig Integral

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

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 thats not the answer. Thank you for your help.

0rthodontist