Evaluate: x ( x 2 + x 2 )^3 d x - Vidyadip

MCQ

Evaluate: $\int \frac{\cos x}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^3} d x$

A
$\frac{2}{\cos \frac{x}{2}+\sin \frac{x}{2}}+C$
B
$\frac{-2}{\cos \frac{x}{2}-\sin \frac{x}{2}}+C$
✓
$\frac{-2}{\cos \frac{x}{2}+\sin \frac{x}{2}}+C$
D
$\frac{2}{\cos \frac{x}{2}-\sin \frac{x}{2}}+C$

✓

Answer

Correct option: C.

$\frac{-2}{\cos \frac{x}{2}+\sin \frac{x}{2}}+C$

(c): We have,
$
\begin{array}{l}
\quad \int \frac{\cos x}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^3} d x=\int \frac{\cos ^2(x / 2)-\sin ^2(x / 2)}{\{\cos (x / 2)+\sin (x / 2)\}^3} d x \\
\text { Put } t=\cos \frac{x}{2}+\sin \frac{x}{2} \Rightarrow 2 d t=\left[\cos \frac{x}{2}-\sin \frac{x}{2}\right] d x \\
\Rightarrow \int \frac{\cos (x / 2)-\sin (x / 2)}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2} d x=2 \int \frac{1}{t^2} d t \\
\quad=\frac{-2}{t}+C=\frac{-2}{\cos (x / 2)+\sin (x / 2)}+C
\end{array}
$

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