# Calculus I: integral

• Mar 28th 2009, 07:36 PM
NonCommAlg
Calculus I: integral
Evaluate $\int \frac{x^2 + 2}{(x \cos x + \sin x)^2} \ dx.$
• Mar 29th 2009, 03:22 AM
halbard
Try the substitution $u=x-\arctan\frac1x$. Then $\tan u=\frac{\tan x-\frac1x}{1+\frac{\tan x}x} =\frac{x\sin x-\cos x}{x\cos x+\sin x}$. Furthermore $1+\tan^2 u=\frac{x^2+1}{(x\cos x+\sin x)^2}$. Also $\mathrm{d}u=\frac{x^2+2}{x^2+1}\mathrm{d}x$. Therefore $\displaystyle\int\dfrac{x^2+2}{(x\cos x+\sin x)^2}\mathrm{d}x=\int(1+\tan^2 u)\mathrm{d}u=\tan u+c=\frac{x\sin x-\cos x}{x\cos x+\sin x}+c$.
• Mar 29th 2009, 02:05 PM
NonCommAlg
Quote:

Originally Posted by halbard

Try the substitution $u=x-\arctan\frac1x$. Then $\tan u=\frac{\tan x-\frac1x}{1+\frac{\tan x}x} =\frac{x\sin x-\cos x}{x\cos x+\sin x}$. Furthermore $1+\tan^2 u=\frac{x^2+1}{(x\cos x+\sin x)^2}$. Also $\mathrm{d}u=\frac{x^2+2}{x^2+1}\mathrm{d}x$. Therefore $\displaystyle\int\dfrac{x^2+2}{(x\cos x+\sin x)^2}\mathrm{d}x=\int(1+\tan^2 u)\mathrm{d}u=\tan u+c=\frac{x\sin x-\cos x}{x\cos x+\sin x}+c$.

correct! well-done! (Clapping)