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Math Help - Calculus I: integral

  1. #1
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    Calculus I: integral

    Evaluate \int \frac{x^2 + 2}{(x \cos x + \sin x)^2} \ dx.
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    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.
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    Quote Originally Posted by halbard View Post

    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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