Q176 / 240 · KCET 2026 · Mathematics
If '$n$' is a natural number, then $\int \dfrac{\sin^n x}{\cos^{n+2} x}\,dx = $
- A$\dfrac{\tan^{n-1} x}{n - 1} + C$
- B$\dfrac{\tan^n x}{n} + C$
- C$\dfrac{\tan^{n+2} x}{n + 2} + C$
- Correct answerD$\dfrac{\tan^{n+1} x}{n + 1} + C$
Solution
The given integral is $I = \int \dfrac{\sin^n x}{\cos^{n+2} x} \,dx$
This can be rewritten as:
$I = \int \left(\dfrac{\sin x}{\cos x}\right)^n \cdot \dfrac{1}{\cos^2 x} \,dx$
$I = \int \tan^n x \sec^2 x \,dx$
Let $\tan x = t$, then $\sec^2 x \,dx = dt$
Substituting these into the integral, we get:
$I = \int t^n \,dt$
$I = \dfrac{t^{n+1}}{n+1} + C$
$I = \dfrac{\tan^{n+1} x}{n+1} + C$
Answer: $\dfrac{\tan^{n+1} x}{n + 1} + C$
This can be rewritten as:
$I = \int \left(\dfrac{\sin x}{\cos x}\right)^n \cdot \dfrac{1}{\cos^2 x} \,dx$
$I = \int \tan^n x \sec^2 x \,dx$
Let $\tan x = t$, then $\sec^2 x \,dx = dt$
Substituting these into the integral, we get:
$I = \int t^n \,dt$
$I = \dfrac{t^{n+1}}{n+1} + C$
$I = \dfrac{\tan^{n+1} x}{n+1} + C$
Answer: $\dfrac{\tan^{n+1} x}{n + 1} + C$
Hence, the correct answer is option: 'D'.