Q89 / 150 · MHT CET 2026 (18 April Shift 1) · Mathematics
The value of $\int\dfrac{n\sqrt{\text{cosec}^2 x^n - 1}}{x^{(1-n)}}\,dx$ is ...
- A$\log(\sin x) + c$
- Correct answerB$\log(\sin x^n) + c$
- C$\log(\cot x^n) + c$
- D$\log(\cos x^n) + c$
Solution
Let the given integral be $I = \int \dfrac{n\sqrt{\text{cosec}^2 x^n - 1}}{x^{(1-n)}} \,dx$
Using the trigonometric identity $\text{cosec}^2 \theta - 1 = \cot^2 \theta$, the integral simplifies to:
$I = \int \dfrac{n \cot x^n}{x^{1-n}} \,dx$
$I = \int n x^{n-1} \cot x^n \,dx$
Let $x^n = t$, which gives $n x^{n-1} \,dx = dt$.
Substituting these into the integral, we get:
$I = \int \cot t \,dt$
$I = \log(\sin t) + c$
Substituting $t = x^n$ back into the expression:
$I = \log(\sin x^n) + c$
Answer: $\log(\sin x^n) + c$
Using the trigonometric identity $\text{cosec}^2 \theta - 1 = \cot^2 \theta$, the integral simplifies to:
$I = \int \dfrac{n \cot x^n}{x^{1-n}} \,dx$
$I = \int n x^{n-1} \cot x^n \,dx$
Let $x^n = t$, which gives $n x^{n-1} \,dx = dt$.
Substituting these into the integral, we get:
$I = \int \cot t \,dt$
$I = \log(\sin t) + c$
Substituting $t = x^n$ back into the expression:
$I = \log(\sin x^n) + c$
Answer: $\log(\sin x^n) + c$
Hence, the correct answer is option: 'B'.