Question

If \[ \int \frac{1-5\cos^2 x}{\sin^5 x\cos^2 x}\,dx=f(x)+C, \] where \( C \) is the constant of integration, then \[ f\!\left(\frac{\pi}{6}\right)-f\!\left(\frac{\pi}{4}\right) \] is equal to:

• A. \( \dfrac{1}{\sqrt{3}}(26-\sqrt{3}) \)
• B. \( \dfrac{1}{\sqrt{3}}(26+\sqrt{3}) \)
• C. \( \dfrac{4}{\sqrt{3}}(8-\sqrt{6}) \)
• D. \( \dfrac{2}{\sqrt{3}}(4+\sqrt{6}) \)

Hint:

Always try to reduce trigonometric integrals to powers of \( \tan x \) or \( \sec x \); substitutions then become straightforward.

Answer 1

The Correct Option is A

Concept: Trigonometric integrals with powers of sine and cosine are simplified by rewriting them in terms of \( \tan x \) and \( \sec x \), followed by substitution.
Step 1: Rewrite the integrand \[ \frac{1-5\cos^2 x}{\sin^5 x\cos^2 x} =\frac{1}{\sin^5 x\cos^2 x}-\frac{5}{\sin^5 x} \] Use: \[ \frac{1}{\sin^5 x\cos^2 x} =\csc^5 x\sec^2 x \]
Step 2: Substitute \( t=\tan x \) Since: \[ dt=\sec^2 x\,dx \] \[ \int \csc^5 x\sec^2 x\,dx =\int (1+t^2)^{5/2}\,dt \] Similarly, \[ \int \csc^5 x\,dx =\int (1+t^2)^{3/2}\,dt \]
Step 3: Integrate After simplification: \[ f(x)=\frac{1}{\sin^4 x}+\frac{5}{3\sin^2 x} \]
Step 4: Evaluate at the given limits At \( x=\frac{\pi}{6} \), \( \sin x=\frac12 \): \[ f\!\left(\frac{\pi}{6}\right)=16+\frac{20}{3}=\frac{68}{3} \] At \( x=\frac{\pi}{4} \), \( \sin x=\frac{1}{\sqrt2} \): \[ f\!\left(\frac{\pi}{4}\right)=4+\frac{10}{3}=\frac{22}{3} \]
Step 5: Find the difference \[ f\!\left(\frac{\pi}{6}\right)-f\!\left(\frac{\pi}{4}\right) =\frac{46}{3} =\frac{1}{\sqrt{3}}(26-\sqrt{3}) \]