Q75 / 180 · COMEDK 2026 (9 May Shift 2) · Mathematics
$\int \left(\sin^6 x + \cos^6 x + 3\sin^2 x \cos^2 x\right)\, dx =$
- A$-\dfrac{3}{2}\cos 2x + C$
- B$\dfrac{2}{3}x + C$
- C$\dfrac{3}{2}\sin 2x + C$
- Correct answerD$x + C$
Solution
Let $I = \int (\sin^6 x + \cos^6 x + 3\sin^2 x \cos^2 x)\, dx$
Using the algebraic identity $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$, we can write:
$\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)^3 - 3\sin^2 x \cos^2 x (\sin^2 x + \cos^2 x)$
Since $\sin^2 x + \cos^2 x = 1$, we get:
$\sin^6 x + \cos^6 x = 1 - 3\sin^2 x \cos^2 x$
Substituting this back into the integrand:
$\sin^6 x + \cos^6 x + 3\sin^2 x \cos^2 x = 1 - 3\sin^2 x \cos^2 x + 3\sin^2 x \cos^2 x = 1$
Therefore, the integral becomes:
$I = \int 1 \, dx = x + C$
Answer: $x + C$
Using the algebraic identity $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$, we can write:
$\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)^3 - 3\sin^2 x \cos^2 x (\sin^2 x + \cos^2 x)$
Since $\sin^2 x + \cos^2 x = 1$, we get:
$\sin^6 x + \cos^6 x = 1 - 3\sin^2 x \cos^2 x$
Substituting this back into the integrand:
$\sin^6 x + \cos^6 x + 3\sin^2 x \cos^2 x = 1 - 3\sin^2 x \cos^2 x + 3\sin^2 x \cos^2 x = 1$
Therefore, the integral becomes:
$I = \int 1 \, dx = x + C$
Answer: $x + C$
Hence, the correct answer is option: 'D'.