Question

The integral of $\sqrt{\sin 2 x} \cos 2 x$ is

• A. $\frac{1}{3}(\sin x)^{3 / 2}+C$
• B. $\frac{1}{3}(\sin x)^{1 / 2}+C$
• C. $\frac{1}{3}(\sin 2 x)^{3 / 2}+C$
• D. $\frac{1}{3}(\sin 2 x)^{1 / 2}+C$

Answer 1

Answer: (C)

$\int \sqrt{\sin 2 x} \cos 2 x d x$
Let $ \sin 2 x=t$
$\Rightarrow 2 \cos 2 x=\frac{d t}{d x}$
$\Rightarrow d x=\frac{d t}{2 \cos 2 x}$
$\therefore \int \sqrt{\sin 2 x} \cos 2 x d x=\int \sqrt{t} \cos 2 x \frac{d t}{2 \cos 2 x}=\frac{1}{2} \int \sqrt{t} d t$
$=\frac{1}{2} \frac{t^{1 / 2+1}}{\left(\frac{1}{2}+1\right)}+C=\frac{1}{3} t^{3 / 2}+C=\frac{1}{3}(\sin 2 x)^{3 / 2}+C$