JEE Mains · Maths · STD 12 - 7.1 indefinite integral
The value of the integral \(\int \frac{\sin \theta \cdot \sin 2 \theta\left(\sin ^{6} \theta+\sin ^{4} \theta+\sin ^{2} \theta\right) \sqrt{2 \sin ^{4} \theta+3 \sin ^{2} \theta+6}}{1-\cos 2 \theta} d \theta\) is (where \(c\) is a constant of integration)
- A \(\frac{1}{18}\left[11-18 \sin ^{2} \theta+9 \sin ^{4} \theta-2 \sin ^{6} \theta\right]^{\frac{3}{2}}+ c\)
- B \(\frac{1}{18}\left[9-2 \cos ^{6} \theta-3 \cos ^{4} \theta-6 \cos ^{2} \theta\right]^{\frac{3}{2}}+c\)
- C \(\frac{1}{18}\left[9-2 \sin ^{6} \theta-3 \sin ^{4} \theta-6 \sin ^{2} \theta\right]^{\frac{3}{2}}+ c\)
- D \(\frac{1}{18}\left[11-18 \cos ^{2} \theta+9 \cos ^{4} \theta-2 \cos ^{6} \theta\right]^{\frac{3}{2}}+ c\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{18}\left[11-18 \cos ^{2} \theta+9 \cos ^{4} \theta-2 \cos ^{6} \theta\right]^{\frac{3}{2}}+ c\)
Step-by-step Solution
Detailed explanation
\(I=\int \frac{\sin \theta \cdot \sin 2 \theta\left(\sin ^{6} \theta+\sin ^{4} \theta+\sin ^{2} \theta\right) \sqrt{2 \sin ^{4} \theta+3 \sin ^{2} \theta+6}}{1-\cos 2 \theta} d \theta\)…
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