Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
The value of $\int_0^{\pi / 2} \frac{d x}{1+\tan ^3 x}$ is
A
$0$
B
1
C
$\frac{\pi}{2}$
✓
$\frac{\pi}{4}$
✓
Answer
Correct option: D.
$\frac{\pi}{4}$
(D) Let $I=\int_0^{\frac{\pi}{2}} \frac{d x}{1+\tan ^3 x}$ $=\int_0^{\frac{\pi}{2}} \frac{\cos ^3 x}{\sin ^3 x+\cos ^3 x} d x$ ...(i) $\therefore \quad I=\int_0^{\frac{\pi}{2}} \frac{\sin ^3 x}{\cos ^3 x+\sin ^3 x} d x$ ...(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 $2 I =\int_0^{\frac{\pi}{2}} d x=[x]_0^{\pi / 2}$ $\therefore \quad 2 I =\frac{\pi}{2} \Rightarrow I =\frac{\pi}{4}$ Alternate Method: $\int_0^{\frac{\pi}{2}} \frac{d x}{1+\tan ^{ n } x}=\int_0^{\frac{\pi}{2}} \frac{d x}{1+\cot ^{ n } x} d x=\frac{\pi}{4}$ $\int_0^{\pi / 2} \frac{d x}{1+\tan ^3 x}=\frac{\pi}{4}$
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