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MHTCET 16 Apr 2026 Shift 1 Paper

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Question : 127 of 150

Marks: +1, -0

\int \cot ^{4} x d x

is equal to

$\frac{\cot ^{3} x}{3}-\cot x+x+c$
$\frac{\cot ^{3} x}{3}+\cot x+x+c$
$-\frac{\cot ^{3} x}{3}+\cot x+x+c$
$\frac{\cot ^{3} x}{3}-2 \cot x+x+c$

Solution:

Concept:

Use the identity

\cot^2 x = \csc^2 x - 1

to break

\cot^4 x

into simpler integrable terms.

Explanation:

Rewrite

\cot^4 x

as

\cot^2 x \cdot \cot^2 x

.

Using the identity, we get:

\cot^4 x = \cot^2 x(\csc^2 x - 1) = \cot^2 x \csc^2 x - \cot^2 x

Therefore,

\int \cot^4 x\,dx = \int \cot^2 x \csc^2 x\,dx - \int \cot^2 x\,dx

For the first integral, let

u = \cot x

, so

du = -\csc^2 x\,dx

.

Thus,

\int \cot^2 x \csc^2 x\,dx = -\int u^2\,du = -\frac{u^3}{3} = -\frac{\cot^3 x}{3}

For the second integral, use the identity again:

\int \cot^2 x\,dx = \int(\csc^2 x - 1)\,dx = -\cot x - x

Combining both results:

\int \cot^4 x\,dx = -\frac{\cot^3 x}{3} - (-\cot x - x) = -\frac{\cot^3 x}{3} + \cot x + x + c

Answer:

-\frac{\cot^{3} x}{3}+\cot x+x+c

Hence, the correct option is C.

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