Question

Evaluate $\int \frac{x^2}{(xsinx+cosx{)}^2}dx$

• A. $\frac{sinx-xcosx}{xsinx+cosx}+c$
• B. $\frac{cosx-xsinx}{xsinx+cosx}+c$
• C. $\frac{cosx-xsinx}{xsinx-cosx}+c$
• D. $\frac{sinx+xcosx}{xsinx+cosx}+c$

Answer 1

The correct option is A $\frac{sinx-xcosx}{xsinx+cosx}+c$

$\int \frac{x^2}{(xsinx+cosx{)}^2}dx$

Put $x=\tan \theta$, $⟹dx={\sec }^2\theta d\theta$

$\int \frac{ta{n}^2\theta se{c}^2\theta }{{\left(\frac{sin\theta }{cos\theta }sin\left(tan\theta \right)+cos\left(tan\theta \right)\right)}^2}d\theta$

$=\int \frac{ta{n}^2\theta }{co{s}^2(tan\theta -\theta )}d\theta$

Substitute $tan\theta -\theta =u\Rightarrow ta{n}^2\theta d\theta =du$

$I=\int \frac{du}{co{s}^{2}u}=\int se{c}^{2}udu=tan(tan\theta -\theta )$

$=tan(x-ta{n}^{-1}x)=\frac{tanx-x}{1+x\tan x}$

$=\frac{sinx-xcosx}{cosx+x\sin x}+C$