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Definite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
$\int_0^{\frac{\pi}{2}} \frac{\cos ^3 x}{\sin x+\cos x} d x$
  • A
    $\frac{\pi-1}{2}$
  • ✓
    $\frac{\pi-1}{4}$
  • C
    $\frac{1+\pi}{4}$
  • D
    $\frac{\pi-3}{4}$
✓

Answer

Correct option: B.
$\frac{\pi-1}{4}$
(B)
Let $I=\int_0^{\frac{\pi}{2}} \frac{\cos ^3 x}{\sin x+\cos x} d x$ …(i)
$I=\int_0^{\frac{\pi}{2}} \frac{\sin ^3 x}{\cos x+\sin x} d x$ ...(ii)
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
Adding (i) and (ii), we get
$2 I =\int_0^{\frac{\pi}{2}} \frac{\sin ^3 x+\cos ^3 x}{\sin x+\cos x} d x$
$\begin{array}{l}=\int_0^{\frac{\pi}{2}}\left(\sin ^2 x-\sin x \cos x+\cos ^2 x\right) d x \\ =\int_0^{\frac{\pi}{2}}(1-\sin x \cos x) d x\end{array}$
$\begin{array}{l}=\int_0^{\frac{\pi}{2}} 1 d x-\int_0^{\frac{\pi}{2}} \sin x \cos x d x \\ =[x]_0^{\pi / 2}-\left[\frac{\sin ^2 x}{2}\right]_0^{\pi / 2}=\frac{\pi}{2}-\frac{1}{2}\end{array}$
$\therefore \quad 2 I=\frac{\pi-1}{2} \Rightarrow I=\frac{\pi-1}{4}$

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