^{-3/7} x · ^{-11/7} x d x=

Correct Answer is : −74tan⁡−47x+c-7/4 ^{-4/7} x+c4−7tan7−4x+c, where c is a constant of integration.

-4/7 ^{-4/7} x+c, where c is a constant of integration.

4/7 ^{4/7} x+c, where c is a constant of integration.

-7/4 ^{-4/7} x+c, where c is a constant of integration.

7/4 ^{4/7} x+c, where c is a constant of integration.

Correct Answer is : −74tan⁡−47x+c-7/4 ^{-4/7} x+c4−7tan7−4x+c, where c is a constant of integration.

Test Index

MHTCET 16 May 2024 Shift 2 Paper

Question : 128 of 150

Marks: +1, -0

\int \cos ^{\frac{-3}{7}} x \cdot \sin ^{\frac{-11}{7}} x d x=

$\frac{-4}{7} \tan ^{\frac{-4}{7}} x+c$ , where $c$ is a constant of integration.
$\frac{4}{7} \tan ^{\frac{4}{7}} x+c$ , where c is a constant of integration.
$\frac{-7}{4} \tan ^{\frac{-4}{7}} x+c$ , where c is a constant of integration.
$\frac{7}{4} \tan ^{\frac{4}{7}} x+c$ , where c is a constant of integration.

Solution:

Concept:

Rewrite the integrand in terms of

\tan x

and

\sec^2 x

, then apply the standard power rule for integration.

Explanation:

We are given:

I = \int \cos^{-\frac{3}{7}} x \cdot \sin^{-\frac{11}{7}} x \, dx

Use the identity

\sin x = \tan x \cdot \cos x

Then:

\sin^{-\frac{11}{7}} x = \cos^{-\frac{11}{7}} x \cdot \tan^{-\frac{11}{7}} x

Multiplying by

\cos^{-\frac{3}{7}} x

gives:

\cos^{-\frac{14}{7}} x \cdot \tan^{-\frac{11}{7}} x = \sec^2 x \cdot \tan^{-\frac{11}{7}} x

So the integral becomes:

I = \int \tan^{-\frac{11}{7}} x \cdot \sec^2 x \, dx

Let

u = \tan x

, so

du = \sec^2 x \, dx

Thus:

I = \int u^{-\frac{11}{7}} \, du = \frac{u^{-\frac{4}{7}}}{-\frac{4}{7}} + c

Simplify:

I = -\frac{7}{4} \tan^{-\frac{4}{7}} x + c

Answer:

\boxed{-\frac{7}{4} \tan^{-\frac{4}{7}} x + c}

Hence, the correct option is C.

Go to Question:

Prev Question

Next Question