Question

Evaluate \[ \int \frac{\sin^2x\tan x} {\cos^6x+\sin^4x\cos^2x+\cos^4x\sin^2x+\sin^6x}\,dx. \]

• A. \[ \log(\sin^4x+\cos^4x)+c \]
• B. \[ \frac14\log(\sin^4x+\cos^4x)+c \]
• C. \[ \frac14\log(1+\tan^4x)+c \]
• D. \[ \log(1+\tan^4x)+c \]

Hint:

Whenever the denominator contains \[ 1+\tan^2x+\tan^4x+\tan^6x, \] use the factorization \[ \boxed{ 1+\tan^2x+\tan^4x+\tan^6x = (1+\tan^2x)(1+\tan^4x), } \] followed by the substitution \[ \boxed{t=\tan x.} \]

Answer 1

The Correct Option is C

Step 1: Simplify the denominator. Factor \[ \cos^6x \] from the denominator: \[ \cos^6x+\sin^4x\cos^2x+\cos^4x\sin^2x+\sin^6x = \cos^6x(1+\tan^2x+\tan^4x+\tan^6x). \] Since \[ 1+\tan^2x+\tan^4x+\tan^6x = (1+\tan^2x)(1+\tan^4x), \] the denominator becomes \[ \cos^6x(1+\tan^2x)(1+\tan^4x). \]

Step 2: Rewrite the numerator. Now, \[ \sin^2x\tan x = \frac{\sin^3x}{\cos x} = \tan^3x\cos^2x. \] Hence, the integrand simplifies to \[ \frac{\tan^3x} {(1+\tan^2x)(1+\tan^4x)}. \] Let \[ t=\tan x. \] Then, \[ dt=(1+t^2)\,dx. \] Therefore, \[ I = \int \frac{t^3}{(1+t^4)(1+t^2)} (1+t^2)\,dx = \int \frac{t^3}{1+t^4}\,dt. \]

Step 3: Integrate. Let \[ u=1+t^4. \] Then, \[ du=4t^3\,dt. \] Hence, \[ I = \frac14 \int \frac{du}{u} = \frac14\log|u|+c. \] Substituting back, \[ I = \frac14 \log(1+\tan^4x)+c. \] Therefore, \[ \boxed{ \frac14\log(1+\tan^4x)+c }. \] Thus, \[ \boxed{(C)} \] is the correct answer.