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

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

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

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

x- 2 x+c_{}, where c is the constant of integration.

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

Test Index

MHTCET 26 Apr 2025 Shift 2 Paper

Question : 121 of 150

Marks: +1, -0

\int \frac{d x}{\sin ^{2} x \cos ^{2} x}=

Solution:

Concept:

Use

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

to rewrite the integrand into standard integrals.

Explanation:

Replace the numerator

1

with

\sin^2 x + \cos^2 x

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

Split the fraction into two separate terms:

= \int \left( \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} \right) dx

= \int \sec^2 x \, dx + \int \csc^2 x \, dx

Use the standard results

\int \sec^2 x \, dx = \tan x

and

\int \csc^2 x \, dx = -\cot x

Therefore:

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

Answer:

Option B:

\tan x - \cot x + c

, where

c

is the constant of integration.

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