\(\int \frac{\tan^{4}\sqrt{x} \sec^{2}\s

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Indefinite Integration

Question:

$\int \frac{\tan^{4}\sqrt{x} \sec^{2}\sqrt{x}dx}{\sqrt{x}}$ equals

Options:

$\frac{2}{5}\tan^{2}\sqrt{x}+C$

$\frac{2}{5}\tan^{5}\sqrt{x}+C$

$\frac{5}{2}\tan^{5}\sqrt{x}+C$

None

Correct Answer:

$\frac{2}{5}\tan^{5}\sqrt{x}+C$

Explanation:

$I=\int \frac{\tan^{4}\sqrt{x} \sec^{2}\sqrt{x}}{\sqrt{x}}dx$

let $y=\tan\sqrt{x}$

$2dy=\frac{\sec^2\sqrt{x}}{\sqrt{x}}dx$

so $I=2\int y^4dy=\frac{2}{5}y^5+C$

$=\frac{2}{5}\tan^{5}\sqrt{x}+C$