Definite Integration
Indefinite Integral via IBP — Evaluating Definite Difference
nta_pyq_2026_jan
Grade 12
Question:
If $\displaystyle\int\frac{1-5\cos^2x}{\sin^5x\cos^2x}\,dx=f(x)+C$, where $C$ is the constant of integration, then $f\!\left(\dfrac{\pi}{6}\right)-f\!\left(\dfrac{\pi}{4}\right)$ is equal to
$\dfrac{1}{\sqrt{3}}(26+\sqrt{3})$
$\dfrac{1}{\sqrt{3}}(26-\sqrt{3})$
$\dfrac{4}{\sqrt{3}}(8-\sqrt{6})$
$\dfrac{2}{\sqrt{3}}(4+\sqrt{6})$
Step-by-Step Solution
Key Concept: $\frac{1-5\cos^2x}{\sin^5x\cos^2x}=\frac{\sec^2x}{\sin^5x}-\frac{5}{\sin^5x}$. By IBP on $\int\frac{\sec^2x}{\sin^5x}dx$: result simplifies to $f(x)=\frac{\tan x}{\sin^5x}$.
$f(x)=\tan x/\sin^5x$. $f(\pi/6)-f(\pi/4)=\dfrac{4}{\sqrt{3}}(8-\sqrt{6})$.
Correct Answer: 3