JEE PYQ: Indefinite Integration Question 5
Question 5 - 2021 (25 Feb Shift 1)
The value of the integral $\int \frac{\sin\theta \cdot \sin 2\theta(\sin^6\theta + \sin^4\theta + \sin^2\theta)\sqrt{2\sin^4\theta + 3\sin^2\theta + 6}}{1 - \cos 2\theta},d\theta$ is (where $c$ is a constant of integration):
(1) $\frac{1}{18}\left[9 - 2\sin^6\theta - 3\sin^4\theta - 6\sin^2\theta\right]^{3/2} + c$
(2) $\frac{1}{18}\left[11 - 18\sin^2\theta + 9\sin^4\theta - 2\sin^6\theta\right]^{3/2} + c$
(3) $\frac{1}{18}\left[11 - 18\cos^2\theta + 9\cos^4\theta - 2\cos^6\theta\right]^{3/2} + c$
(4) $\frac{1}{18}\left[9 - 2\cos^6\theta - 3\cos^4\theta - 6\cos^2\theta\right]^{3/2} + c$
Show Answer
Answer: (3)
Solution
Let $\sin\theta = t$, $\cos\theta,d\theta = dt$. The integral simplifies to $\frac{1}{12}\int \sqrt{2z},dz = \frac{1}{18}z^{3/2} + c$ where $z = 2t^6 + 3t^4 + 6t^2$. Expressing in terms of $\cos\theta$: $\frac{1}{18}\left[(1-\cos^2\theta)(2\cos^4\theta - 7\cos^2\theta + 11)\right]^{3/2} + c = \frac{1}{18}[11 - 18\cos^2\theta + 9\cos^4\theta - 2\cos^6\theta]^{3/2} + c$.
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