Question

Evaluate: \[ \int \frac{\sin^2 x \cos^2 x}{\cos^6 x+\sin^6 x}\,dx \]

• A. \(\dfrac12\tan^{-1}(\tan^2x)+c\)
• B. \(\dfrac13\tan^{-1}(\tan^2x)+c\)
• C. \(\dfrac13\tan^{-1}(\tan^3x)+c\)
• D. \(\tan^{-1}(\tan^3x)+c\)

Hint:

Whenever expressions contain symmetric powers of \(\sin x\) and \(\cos x\), converting into \(\tan x\) often reduces the integral to a rational form.

Answer 1

The Correct Option is C

Concept: Expressions involving powers of \(\sin x\) and \(\cos x\) are often simplified by converting everything into \(\tan x\). Useful identity: \[ a^3+b^3=(a+b)(a^2-ab+b^2) \] Applying this identity to: \[ \sin^6x+\cos^6x \] leads to major simplification.

Step 1: Simplify the denominator. \[ \sin^6x+\cos^6x = (\sin^2x)^3+(\cos^2x)^3 \] Using: \[ a^3+b^3=(a+b)(a^2-ab+b^2) \] we get: \[ =(\sin^2x+\cos^2x)(\sin^4x-\sin^2x\cos^2x+\cos^4x) \] Since, \[ \sin^2x+\cos^2x=1 \] therefore, \[ \sin^6x+\cos^6x = \sin^4x-\sin^2x\cos^2x+\cos^4x \] Now use: \[ \sin^4x+\cos^4x = (\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x \] \[ =1-2\sin^2x\cos^2x \] Hence, \[ \sin^6x+\cos^6x = 1-3\sin^2x\cos^2x \] Thus the integral becomes: \[ I= \int \frac{\sin^2x\cos^2x} {1-3\sin^2x\cos^2x} dx \]

Step 2: Convert into tangent form. Put: \[ t=\tan x \] Then, \[ dx=\frac{dt}{1+t^2} \] Also, \[ \sin^2x\cos^2x = \frac{t^2}{(1+t^2)^2} \] Substituting carefully and simplifying yields: \[ I= \int \frac{t^2}{1+t^6}\,dt \]

Step 3: Use substitution. Let, \[ u=t^3 \] Then, \[ du=3t^2dt \] or, \[ t^2dt=\frac{du}{3} \] Thus, \[ I= \frac13 \int \frac{du}{1+u^2} \] \[ = \frac13\tan^{-1}u+c \] Substituting back: \[ I= \frac13\tan^{-1}(t^3)+c \] Since \(t=\tan x\), \[ \boxed{ \frac13\tan^{-1}(\tan^3x)+c } \]