Question

$\int \frac{\tan ^4 \sqrt{x} \cdot \sec ^2 \sqrt{x}}{\sqrt{x}} d x=$

• A. $\frac{-5}{2}[\tan \sqrt{x}]^5+C$
• B. $[\tan \sqrt{x}]^5+C$
• C. $\frac{2}{5}[\tan \sqrt{x}]^5+C$
• D. $\frac{5}{2}[\tan \sqrt{x}]^5+C$

Answer 1

Answer: (C)

Let $I=\int \frac{\tan ^4 \sqrt{x} \cdot \sec ^2 \sqrt{x}}{\sqrt{x}} d x$
Put $\tan \sqrt{x}=t$
$ \Rightarrow \frac{\sec ^2 \sqrt{x}}{2 \sqrt{x}} d x=d t$
$\therefore I=\int \frac{\tan ^4 \sqrt{x} \cdot \sec ^2 \sqrt{x}}{\sqrt{x}} d x=\int 2 t^4 d t $
$=\frac{2 t^5}{5}+C=\frac{2(\tan \sqrt{x})^5}{5}+C$