The integral

Question:

The integral $\int \sec ^{2 / 3} x \operatorname{cosec}^{4 / 3} x d x$ is equal to:

  1. (1) $-3 \tan ^{-1 / 3} x+C$

  2. (2) $-\frac{3}{4} \tan ^{-4 / 3} x+C$

  3. (3) $-3 \cot ^{-1 / 3} x+C$

  4. (4) $3 \tan ^{-1 / 3} x+C$


Correct Option: 1

Solution:

$I=\int \sec ^{\frac{2}{3}} x \cdot \operatorname{cosec}^{\frac{4}{3}} d x$

$I=\int \frac{\sec ^{2} x d x}{\tan ^{\frac{4}{3}} x}$

Put $\tan x=z$

$\Rightarrow \sec ^{2} x d x=d z$

$\Rightarrow I=\int z^{-\frac{4}{3}} \cdot d z=\frac{z^{\frac{-1}{3}}}{\left(\frac{-1}{3}\right)}+C \Rightarrow I=-3(\tan x)^{\frac{-1}{3}}+C$

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