Question

\[ \int(\sin^6x+\cos^6x+3\sin^2x\cos^2x)\,dx= \]

• A. \(-\dfrac{3}{2}\cos2x+C\)
• B. \(x+C\)
• C. \(\dfrac{3}{2}\sin2x+C\)
• D. \(\dfrac{2}{3}x+C\)

Hint:

Remember: \[ \sin^6x+\cos^6x=1-3\sin^2x\cos^2x \] This identity is very useful in integration problems.

Answer 1

The Correct Option is B

Understanding the Concept: Before integrating trigonometric expressions, simplify them using identities. Here we use: \[ a^3+b^3=(a+b)^3-3ab(a+b) \] along with: \[ \sin^2x+\cos^2x=1 \]
Step 1: Rewriting the powers. Let: \[ a=\sin^2x,\quad b=\cos^2x \] Then: \[ \sin^6x=a^3,\quad \cos^6x=b^3 \] Thus the integrand becomes: \[ a^3+b^3+3ab \] Since: \[ a+b=\sin^2x+\cos^2x=1 \]
Step 2: Applying the identity. Using: \[ (a+b)^3=a^3+b^3+3ab(a+b) \] and \(a+b=1\), we get: \[ 1=a^3+b^3+3ab \] Hence: \[ \sin^6x+\cos^6x+3\sin^2x\cos^2x=1 \]
Step 3: Integrating. Therefore: \[ \int1\,dx=x+C \]