Evaluate x^2 d x (x x+ x)^2 . - Vidyadip

Question

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

✓

Answer

Let
$
I=\int \frac{x^2 d x}{(x \sin x+\cos x)^2}
$
Since $\frac{d}{d x}(x \sin x+\cos x)=1 \cdot \sin x+x \cos x-\sin x$
$
=x \cos x
$
So assuming $x^2$ as $\frac{x}{\cos x} \cdot x \cos x$ taking
$\frac{x \cos x}{(x \sin x+\cos x)^2}$ as the second function, integrating by parts,
$
\begin{array}{l}
\int \frac{x^2}{(x \sin x+\cos x)^2} d x \\
=\int \frac{x}{\cos x} \cdot \frac{x \cos x}{(x \sin x+\cos x)^2} d x \\
=\frac{x}{\cos x} \times\left(\frac{-1}{x \sin x+\cos x}\right)- \\
\int \frac{\cos x+x \sin x}{\cos ^2 x} \times\left\{\frac{-1}{x \sin x+\cos x}\right\} d x \\
=\frac{-x}{\cos x(x \sin x+\cos x)}+\int \sec ^2 x d x
\end{array}
$
$\begin{array}{l}=\frac{-x}{\cos x(x \sin x+\cos x)}+\tan x+ C \\ =\frac{-x}{\cos x(x \sin x+\cos x)}+\frac{\sin x}{\cos x}+ C \\ =\frac{-x+x \sin ^2 x+\sin x \cos x}{\cos x(x \sin x+\cos x)}+ C \end{array}$
$\begin{array}{l}=\frac{-x \cos ^2 x+\sin x \cos x}{\cos x(x \sin x+\cos x)}+ C \\ =\frac{\sin x-x \cos x}{x \sin x+\cos x}+ C \text { }\end{array}$

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