Question

The value of \(\int \frac{sin^6x+cos^6x}{sin^2x\cdot cos^2x}\,\text{d}x\) is

• A. \(cotx+cot^2x+c\), where c is the constant of integration
• B. \(tanx-cotx-3x+c\), where c is the constant of integration
• C. \(2sin^5x+2cos^5x+c\), where c is the constant of integration
• D. \(sin^3x+2cos^3x+c\), where c is the constant of integration

Hint:

Write sin^6 + cos^6 = 1 - 3 sin^2 cos^2 and split the fraction.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Use the identity \(a^3 + b^3 = (a + b)^3 - 3ab(a + b)\) with \(a = \sin^2x\) and \(b = \cos^2x\). Since \(a + b = 1\), we get \(\sin^6x + \cos^6x = 1 - 3\sin^2x\cos^2x\).

Step 2: Split the integrand
\[ \frac{1 - 3\sin^2x\cos^2x}{\sin^2x\cos^2x} = \frac{1}{\sin^2x\cos^2x} - 3 \]
And \(\dfrac{1}{\sin^2x\cos^2x} = \dfrac{\sin^2x + \cos^2x}{\sin^2x\cos^2x} = \sec^2x + \text{cosec}^2x\).

Step 3: Integrate
\[ \int(\sec^2x + \text{cosec}^2x - 3)\,dx = \tan x - \cot x - 3x + c \]
This is option (B).

Final Answer:
The integral is \(\tan x - \cot x - 3x + c\), option (B). \[ \boxed{\tan x - \cot x - 3x + c} \]