MCQ
$\int\limits_0^{\frac{\pi }{4}} {} (cos 2x)^{3/2}. \cos\, x \,dx =$
  • A
    $\frac{{3\,\pi }}{{16}}$
  • B
    $\frac{{3\,\pi }}{{32}}$
  • $\frac{{3\,\pi }}{{16\,\sqrt 2 }}$
  • D
    $\frac{{3\,\pi \,\sqrt 2 }}{{16}}$

Answer

Correct option: C.
$\frac{{3\,\pi }}{{16\,\sqrt 2 }}$
c
$I =\int\limits_0^{\frac{\pi }{4}} {} (1 - 2\, \sin^2\, x)^{3/2} \,\cos\, x \,dx$.

Put $\sqrt 2\, \sin\, x = \sin \,\theta$
$\Rightarrow I = \frac{1}{{\sqrt 2 }}\,  \int\limits_0^{\frac{\pi }{2}} \,cos^4\, \theta \,d\theta = \frac{{3\,\pi }}{{16\,\sqrt 2 }}$

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