Calculus I: integral

**NonCommAlg** · March 28th 2009, 06:36 PM

Evaluate $\int \frac{x^2 + 2}{(x \cos x + \sin x)^2} \ dx.$

**halbard** · March 29th 2009, 02:22 AM

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 $\displaystyle\int\dfrac{x^2+2}{(x\cos x+\sin x)^2}\mathrm{d}x=\int(1+\tan^2 u)\mathrm{d}u=\tan u+c=\frac{x\sin x-\cos x}{x\cos x+\sin x}+c$ .

**NonCommAlg** · March 29th 2009, 01:05 PM

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!

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