The value of { ^{6} x+ ^{6} x}{ ² x · ² x} d x is

Correct Answer is : tan⁡x−cot⁡x−3x+c x- x-3 x+ctanx−cotx−3x+c, where c is the constant of integration

x+ ² x+c, where c is the constant of integration

x- x-3 x+c, where c is the constant of integration

2 ^{5} x+2 ^{5} x+c, where c is the constant of integration

³ x+2 ³ x+c, where c is the constant of integration

Correct Answer is : tan⁡x−cot⁡x−3x+c x- x-3 x+ctanx−cotx−3x+c, where c is the constant of integration

Test Index

MHTCET 15 Apr 2026 Shift 2 Paper

Question : 129 of 150

Marks: +1, -0

The value of

\int \frac{\sin ^{6} x+\cos ^{6} x}{\sin ^{2} x \cdot \cos ^{2} x} d x

Solution:

Concept:

Use algebraic identities and standard trigonometric integrals to simplify the integrand.

Explanation:

Given integral is

\int \frac{\sin^{6}x + \cos^{6}x}{\sin^{2}x \cos^{2}x} \, dx

Use the identity

a^{3} + b^{3} = (a+b)^{3} - 3ab(a+b)

Take

a = \sin^{2}x

and

b = \cos^{2}x

Then

\sin^{6}x + \cos^{6}x = (\sin^{2}x + \cos^{2}x)^{3} - 3\sin^{2}x\cos^{2}x(\sin^{2}x + \cos^{2}x)

Since

\sin^{2}x + \cos^{2}x = 1

, we get

\sin^{6}x + \cos^{6}x = 1 - 3\sin^{2}x\cos^{2}x

Divide by

\sin^{2}x\cos^{2}x

\frac{\sin^{6}x + \cos^{6}x}{\sin^{2}x\cos^{2}x} = \frac{1}{\sin^{2}x\cos^{2}x} - 3

Also,

\frac{1}{\sin^{2}x\cos^{2}x} = \sec^{2}x + \csc^{2}x

So the integral becomes

\int \left( \sec^{2}x + \csc^{2}x - 3 \right) dx

Integrate term by term:

\int \sec^{2}x \, dx = \tan x, \quad \int \csc^{2}x \, dx = -\cot x, \quad \int 3 \, dx = 3x

Thus,

\int \frac{\sin^{6}x + \cos^{6}x}{\sin^{2}x\cos^{2}x} dx = \tan x - \cot x - 3x + c

Answer:

\tan x - \cot x - 3x + c

, hence Option B.

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