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Definite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
$\int_0^{\frac{\pi}{2}} \frac{\sin ^3 x \cos x d x}{\sin ^4 x+\cos ^4 x}=$
  • A
    $\pi$
  • B
    $\frac{\pi}{2}$
  • C
    $\frac{\pi}{4}$
  • ✓
    $\frac{\pi}{8}$
✓

Answer

Correct option: D.
$\frac{\pi}{8}$
(D)
Let $I =\int_0^{\frac{\pi}{2}} \frac{\sin ^3 x \cos x}{\sin ^4 x+\cos ^4 x} d x$
$=\int_0^{\frac{\pi}{2}} \frac{\sin ^3\left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)}{\sin ^4\left(\frac{\pi}{2}-x\right)+\cos ^4\left(\frac{\pi}{2}-x\right)} d x$
$=\int_0^{\frac{\pi}{2}} \frac{\cos ^3 x \sin x}{\cos ^4 x+\sin ^4 x} d x$
$\therefore \quad 2 I =\int_0^{\frac{\pi}{2}} \frac{\sin ^3 x \cos x+\cos ^3 x \sin x}{\cos ^4 x+\sin ^4 x} d x$
$=\int_0^{\frac{\pi}{2}} \frac{\sin x \cos x}{\cos ^4 x\left(1+\tan ^4 x\right)} d x$
$=\int_0^{\frac{\pi}{2}} \frac{\tan x \sec ^2 x}{\left(1+\tan ^4 x\right)} d x$
Put $\tan ^2 x= t$
$\therefore \quad \tan x \sec ^2 x d x=\frac{ dt }{2}$
When $x=0, t =0$ and when $x=\frac{\pi}{2}, t =\infty$
$\therefore \quad 2 I =\frac{1}{2} \int_0^{\infty} \frac{ dt }{1+ t ^2}$
$\therefore \quad I=\frac{1}{4}\left[\tan ^{-1} t\right]_0^{\infty}$
$\begin{array}{l}=\frac{1}{4}\left(\frac{\pi}{2}-0\right) \\ =\frac{\pi}{8}\end{array}$

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