Q79 / 150 · MHT CET 2026 (16 April Shift 1) · Mathematics
$\displaystyle\int \cot^4 x\,dx$ is equal to
- A$\dfrac{\cot^3 x}{3} - \cot x + x + c$
- B$\dfrac{\cot^3 x}{3} + \cot x + x + c$
- Correct answerC$-\dfrac{\cot^3 x}{3} + \cot x + x + c$
- D$\dfrac{\cot^3 x}{3} - 2\cot x + x + c$
Solution
$I = \int \cot^4 x \, dx$
$I = \int \cot^2 x (\csc^2 x - 1) \, dx$
$I = \int \cot^2 x \csc^2 x \, dx - \int \cot^2 x \, dx$
$I = \int \cot^2 x \csc^2 x \, dx - \int (\csc^2 x - 1) \, dx$
Substituting $\cot x = t \Rightarrow -\csc^2 x \, dx = dt$ in the first integral:
$I = -\int t^2 \, dt - (-\cot x - x) + c$
$I = -\dfrac{t^3}{3} + \cot x + x + c$
$I = -\dfrac{\cot^3 x}{3} + \cot x + x + c$
Answer: $-\dfrac{\cot^3 x}{3} + \cot x + x + c$
$I = \int \cot^2 x (\csc^2 x - 1) \, dx$
$I = \int \cot^2 x \csc^2 x \, dx - \int \cot^2 x \, dx$
$I = \int \cot^2 x \csc^2 x \, dx - \int (\csc^2 x - 1) \, dx$
Substituting $\cot x = t \Rightarrow -\csc^2 x \, dx = dt$ in the first integral:
$I = -\int t^2 \, dt - (-\cot x - x) + c$
$I = -\dfrac{t^3}{3} + \cot x + x + c$
$I = -\dfrac{\cot^3 x}{3} + \cot x + x + c$
Answer: $-\dfrac{\cot^3 x}{3} + \cot x + x + c$
Hence, the correct answer is option: 'C'.