Question

Evaluate \[ \int \frac{\cos^4x}{\left(\sin^2x+\sin^{-3}x\cos^5x\right)^3}\,dx \]

• A. \(\frac{1}{5}(1+\cot^5x)^{-2}+C\)
• B. \(\frac{1}{10}(1+\cot^2x)^{-5}+C\)
• C. \(\frac{1}{10}(1+\cot^5x)^{-2}+C\)
• D. \(\frac{1}{5}(1+\cot^5x)^{-5}+C\)

Hint:

When an integrand contains powers of \(\cot x\) and \(\cosec^2x\), try the substitution \(u=1+\cot^n x\).

Answer 1

The Correct Option is C

Step 1: Simplify the denominator.
Given denominator term is \[ \sin^2x+\sin^{-3}x\cos^5x \] Now, \[ \sin^{-3}x\cos^5x=\frac{\cos^5x}{\sin^3x} \] So, \[ \sin^2x+\frac{\cos^5x}{\sin^3x} = \sin^2x\left(1+\frac{\cos^5x}{\sin^5x}\right) \] \[ =\sin^2x(1+\cot^5x) \] Therefore, \[ \left(\sin^2x+\sin^{-3}x\cos^5x\right)^3 = \sin^6x(1+\cot^5x)^3 \]

Step 2: Rewrite the integral.
The integral becomes \[ \int \frac{\cos^4x}{\sin^6x(1+\cot^5x)^3}\,dx \] Now, \[ \frac{\cos^4x}{\sin^6x} = \cot^4x\cosec^2x \] Thus, \[ \int \frac{\cos^4x}{\left(\sin^2x+\sin^{-3}x\cos^5x\right)^3}\,dx = \int \frac{\cot^4x\cosec^2x}{(1+\cot^5x)^3}\,dx \]

Step 3: Use substitution.
Let \[ u=1+\cot^5x \] Then, \[ \frac{du}{dx}=5\cot^4x(-\cosec^2x) \] \[ du=-5\cot^4x\cosec^2x\,dx \] So, \[ \cot^4x\cosec^2x\,dx=-\frac{du}{5} \]

Step 4: Integrate.
Therefore, \[ \int \frac{\cot^4x\cosec^2x}{(1+\cot^5x)^3}\,dx = -\frac{1}{5}\int u^{-3}\,du \] \[ =-\frac{1}{5}\cdot \frac{u^{-2}}{-2} \] \[ =\frac{1}{10}u^{-2}+C \] Substituting back, \[ =\frac{1}{10}(1+\cot^5x)^{-2}+C \]

Step 5: Final conclusion.
Hence, \[ \boxed{\frac{1}{10}(1+\cot^5x)^{-2}+C} \]