Solve this following

Question:

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

(Hence $\mathrm{C}$ is a constant of integration)

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

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

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

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


Correct Option: , 4

Solution:

$I=\int \frac{d x}{(\sin x)^{4 / 3} \cdot(\cos x)^{2 / 3}}$

$I=\int \frac{d x}{\left(\frac{\sin x}{\cos x}\right)^{4 / 3} \cdot \cos ^{2} x}$

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

put $\tan x=t \Rightarrow \sec ^{2} x d x=d t$

$\therefore \mathrm{I}=\int \frac{\mathrm{dt}}{\mathrm{t}^{4 / 3}} \Rightarrow \mathrm{I}=\frac{-3}{\mathrm{t}^{1 / 3}}+\mathrm{c}$

$\Rightarrow I=\frac{-3}{(\tan x)^{1 / 3}}+c$

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