Question

The value of the integral ∫ sinθ sin2θ (sin⁶θ + sin⁴θ + sin²θ) / √(2sin⁴θ + 3sin²θ + 6) dθ is: (where c is a constant of integration)

• A. (1/18) [9 - 2sin⁶θ - 3sin⁴θ - 6sin²θ]\^{3/2} + c
• B. (1/18) [9 - 2cos⁶θ - 3cos⁴θ - 6cos²θ]\^{3/2} + c
• C. (1/18) [11 - 18sin²θ + 9sin⁴θ - 2sin⁶θ]\^{3/2} + c
• D. (1/18) [11 - 18cos²θ + 9cos⁴θ - 2cos\^6θ]\^{3/2} + c

Hint:

For complex trigonometric integrals, look for a substitution where the derivative of the expression inside the root matches the numerator.

Answer 1

The Correct Option is D

Step 1: Simplify the numerator: $\sin \theta (2\sin \theta \cos \theta) (\sin^6 \theta + \sin^4 \theta + \sin^2 \theta) = 2\sin^2 \theta \cos \theta (.......)$.
Step 2: Substitute $x = \sin^2 \theta$, then $dx = 2\sin \theta \cos \theta d\theta$. The integral structure suggests converting everything to $\cos \theta$. By substituting $t = \cos^2 \theta$ and carefully evaluating the differential, we reach the expression under the square root in (D).