Question #73167

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Question #73167

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Answer 1 · 2017-12-12T05:02:00

(sinx -xcosx)/(xsinx+cosx)+C

Explanation:

$\int \frac{x^{2} d x}{{(x sin x + cos x)}^{2}} = \int - \frac{x}{cos x} d (\frac{1}{x sin x + cos x})$ $int (x^2 dx)/((x sin x +cos x)^2)= int -x/(cos x) d(1/(xsinx+cosx))$

$= \frac{- x sec x}{x sin x + cos x} + \int \frac{sec x + x sec x tan x}{x sin x + cos x} d x$ $=(-xsecx)/(xsin x+cos x)+int (sec x + xsecx tanx)/(xsinx+cosx) dx$

But
$\frac{sec x + x sec x tan x}{x sin x + cos x} = \frac{\frac{1}{cos x} + \frac{x sin x}{{cos}^{2} x}}{x sin x + cos x}$ $(sec x + xsecx tanx)/(xsinx+cosx) =(1/(cos x) + (xsinx)/(cos^2x))/(xsin x+ cosx)$

= $1/cos^2x\cancel (xsinx +cosx)/\cancel(xsinx+cosx)$
= $sec^2x$

then
$int (x^2 dx)/((x sin x +cos x)^2) = (-xsecx)/(xsin x+cos x) + int sec^2x dx$

$= (-xsecx)/(xsin x+cos x)+tanx + C$
$= (xsin^2x + sin xcos x - x)/(cos x(xsin x+cos x))+C$
$=(sinx -xcosx)/(xsinx+cosx)+C$

We can verify by differentiating

$f'(x) = ((xsin x+cos x)(\cancel(cosx)-\cancel(cosx)+xsinx) - (sinx - xcosx)(xcosx + \cancel (sinx) - cancel(sinx)))/(xsinx+cosx)^2$

$=(x^2sin^2x+\cancel(xsinxcosx)-\cancel(xsinxcosx)+x^2cos^2x)/(xsinx+cosx)^2$
= $x^2/(xsinx+cosx)^2$

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Question #73167

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