MCQ
$\int\limits_{ - 3\pi }^{3\pi } {{{\sin }^2}\theta \,{{\sin }^2}\,2\theta d\theta }$ is equal to -
- A$\pi$
- ✓$\frac{3 \pi}{2}$
- C$\frac{5 \pi}{2}$
- D$6 \pi$
$=8 \int_{0}^{3 \pi} \sin ^{4} \theta \cos ^{2} \theta \mathrm{d} \theta=24 \int_{0}^{\pi} \sin ^{4} \theta \cos ^{2} \theta \mathrm{d} \theta$
$=48 \int_{0}^{\pi / 2} \sin ^{4} \theta \cos ^{2} \theta d \theta$
$=\frac{48 \cdot(3.1) \cdot(1)}{6.4 \cdot 2} \cdot \frac{\pi}{2}=\frac{3 \pi}{2}$
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