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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
MCQ
Evaluate: $\int \sin ^3 x \cos ^3 x d x$
  • A
    $\frac{-1}{32}\left\{\frac{-3}{2} \cos 2 x+\frac{1}{6} \cos 6 x\right\}+C$
  • B
    $\frac{1}{32}\left\{\frac{-3}{2} \cos 6 x+\frac{1}{6} \cos 2 x\right\}+C$
  • ✓
    $\frac{1}{32}\left\{\frac{-3}{2} \cos 2 x+\frac{1}{6} \cos 6 x\right\}+C$
  • D
    None of these
✓

Answer

Correct option: C.
$\frac{1}{32}\left\{\frac{-3}{2} \cos 2 x+\frac{1}{6} \cos 6 x\right\}+C$
Let $I=\int \sin ^3 x \cos ^3 x d x$
$\Rightarrow I=\frac{1}{8} \int(2 \sin x \cos x)^3 d x$
$\Rightarrow I=\frac{1}{8} \int \sin ^3 2 x d x $
$\Rightarrow I=\frac{1}{8} \int \frac{3 \sin 2 x-\sin 6 x}{4} d x$
$\Rightarrow I=\frac{1}{32}\left\{-\frac{3}{2} \cos 2 x+\frac{1}{6} \cos 6 x\right\}+C$

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