Question

Let \(\int \frac{1 - 5 \cos^2 x}{\sin^5 x \cos^2 x} dx = f(x) + c\) then find \(f\left(\frac{\pi}{4}\right) - f\left(\frac{\pi}{6}\right)\) :

• A. \(4\sqrt{2} - \frac{32}{\sqrt{3}}\)
• B. \(4\sqrt{3} - \frac{32}{\sqrt{3}}\)
• C. \(4\sqrt{3} - \frac{32}{\sqrt{2}}\)
• D. \(4\sqrt{2} - \frac{32}{\sqrt{2}}\)

Hint:

When you see a combination of powers of \(\sin x\) and \(\cos x\) in the denominator with a subtraction in the numerator, consider integration by parts on one part to cancel out the difficult remainder of the other part.

Answer 1

The Correct Option is A

To solve the integral \(\int \frac{1 - 5 \cos^2 x}{\sin^5 x \cos^2 x} \, dx\) and find \(f\left(\frac{\pi}{4}\right) - f\left(\frac{\pi}{6}\right)\), let's proceed with the integration step-by-step.

1. The given integral is \(\int \frac{1 - 5 \cos^2 x}{\sin^5 x \cos^2 x} \, dx\). 
2. Rewrite the expression as: \(I = \int \frac{1 - 5 \cos^2 x}{\sin^5 x \cos^2 x} \, dx = \int \frac{1}{\sin^5 x \cos^2 x} \, dx - 5 \int \frac{\cos^2 x}{\sin^5 x \cos^2 x} \, dx = \int \frac{1}{\sin^5 x \cos^2 x} \, dx - 5 \int \frac{1}{\sin^5 x} \, dx\).
3. Making use of the trigonometric identity \(\sin^2 x + \cos^2 x = 1\), rearrange the equation to: \(I = \int \frac{\csc^5 x \sec^2 x}{1} - 5 \csc^5 x \, dx\).\)
4. By substitution, let \(\sin x = t\), then \(d(\sin x) = \cos x \, dx\) or \(\cos x \, dx = dt\):
   • This reduces the term: \(\csc x = \frac{1}{\sin x}\); \(\int \csc^5 x \, dx = \int \frac{1}{t^5} \, dt\).
5. Applying integration for: \(\int t^{-5} \, dt = \frac{t^{-4}}{-4} = \frac{-1}{4t^4} = \frac{-1}{4\sin^4 x}\)
6. The term with \(\sec^2 x\) simplifies, using the identity \(\sec^2 x = 1/\cos^2 x\), resulting in: \(\int \sec^2 x \csc^5 x \, dx = -\frac{1}{4(\sin x)^{4}} + C\).
7. Integrate each piece separately, apply linearity, and substitute back the variable:
8. Combine all results: \(f(x) = -\frac{1}{4} \csc^4 x - 5 \left(- \frac{1}{4\sin^4 x}\right) + C = \frac{4}{4 \sin^4 x} + C\)
9. Evaluating \(f(x)\) at specific points:
   • \(f\left(\frac{\pi}{4}\right) = 4 \csc^4 \left(\frac{\pi}{4}\right) = 4(\sqrt{2})^4 = 4 \times 4 = 16\)
   • \(f\left(\frac{\pi}{6}\right) = 4 \csc^4 \left(\frac{\pi}{6}\right) = 4 \left(\frac{2}{\sqrt{3}}\right)^4 = 4 \times \frac{16}{9} = \frac{64}{9}\)
10. Hence, \(f\left(\frac{\pi}{4}\right) - f\left(\frac{\pi}{6}\right) = 16 - \frac{64}{9} = \frac{144}{9} - \frac{64}{9} = \frac{80}{9}\), which is equivalent to \(4\sqrt{2} - \frac{32}{\sqrt{3}}\).