²/3} x · {cosec}^{4/3} x d x=

Correct Answer is : −3tan⁡−13x+c′-3 ^{-1/3} x+c^{}−3tan3−1x+c′ where ccc is the constant of integration

3 ^{-1/3} x+c^{}, where c is the constant of integration

-3 ^{-1/3} x+c^{} where c is the constant of integration

-3/4 ^{-4/3} x+c, where c is the constant of integration Let I= ²/3} x {cosec}^{4/3} x d x {aligned}& = {d x}{ ^{4/3} x ²/3} x} \\& = {d x}{≤ft( x/ x)^{4/3} ² x} \\& = { ² x}{ ^{4/3} x} d x{aligned} Put x=t ² x d x=d t {aligned}& ∴ I= {d t}{t^{4/3}}={-3}{t¹/3}}+c \\& =-3 ^{-1/3} x+c{aligned}

²/3} x · {cosec}^{4/3} x d x=

Correct Answer is : −3tan⁡−13x+c′-3 ^{-1/3} x+c^{}−3tan3−1x+c′ where ccc is the constant of integration

Test Index

MHTCET 22 Apr 2025 Shift 2 Paper

Question : 138 of 150

Marks: +1, -0

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

$3 \tan ^{\frac{-1}{3}} x+c^{\prime}$ , where c is the constant of integration
$-3 \tan ^{\frac{-1}{3}} x+c^{\prime}$ where $c$ is the constant of integration
$-3 \cot ^{\frac{-1}{3}} x+c^{\prime}$ where $c$ is the constant of integration
$-\frac{3}{4} \tan ^{\frac{-4}{3}} x+c$ , where c is the constant of integration
Let $\quad I=\int \sec ^{\frac{2}{3}} x \operatorname{cosec}^{\frac{4}{3}} x d x$
$\begin{aligned}& =\int \frac{d x}{\sin ^{\frac{4}{3}} x \cos ^{\frac{2}{3}} x} \\& =\int \frac{d x}{\left(\frac{\sin x}{\cos x}\right)^{\frac{4}{3}} \cos ^{2} x} \\& =\int \frac{\sec ^{2} x}{\tan ^{\frac{4}{3}} x} d x\end{aligned}$
Put $\quad \tan x=t \Rightarrow \sec ^{2} x d x=d t$
$\begin{aligned}& \therefore \quad I=\int \frac{d t}{t^{\frac{4}{3}}}=\frac{-3}{t^{\frac{1}{3}}}+c \\& =-3 \tan ^{\frac{-1}{3}} x+c\end{aligned}$

Validate

Solution:

Concept:

Rewrite the integrand in terms of

\tan x

and use substitution

t = \tan x

Explanation:

Let

I = \int \sec^{2/3} x \cdot \csc^{4/3} x \, dx

Rewrite as

I = \int \frac{dx}{\sin^{4/3} x \cos^{2/3} x}

Using

\sin x = \tan x \cos x

I = \int \frac{dx}{\tan^{4/3} x \cdot \cos^2 x}

So,

I = \int \frac{\sec^2 x}{\tan^{4/3} x} \, dx

Put

\tan x = t

, so

\sec^2 x \, dx = dt

I = \int t^{-4/3} \, dt = \frac{t^{-1/3}}{-1/3} + c

I = -3 \tan^{-1/3} x + c

Answer:

-3 \tan^{-1/3} x + c

Hence, the correct option is B.

Go to Question:

Prev Question

Next Question