Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
$\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\cot x}}$ is
A
$\frac{\pi}{3}$
B
$\frac{\pi}{6}$
✓
$\frac{\pi}{12}$
D
$\frac{\pi}{2}$
✓
Answer
Correct option: C.
$\frac{\pi}{12}$
(C) Let $I =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{1+\sqrt{\cot x}}$ $=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} dx$ ...(i) $\therefore \quad I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x$ ...(ii) $\ldots\left[\because \int_{ a }^{ b } f (x) d x=\int_{ a }^{ b } f ( a + b -x) d x\right]$ Adding (i) and (ii), we get $2 I =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} d x$ $\therefore \quad I =\frac{1}{2}\left(\frac{\pi}{3}-\frac{\pi}{6}\right)=\frac{\pi}{12}$ Alternate Method: $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{1+\sqrt{\cot x}}=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$ Here $f (x)=\sqrt{\sin x}, a =\frac{\pi}{6}$ and $b =\frac{\pi}{3}$ $\int_{ a }^{ b } \frac{ f (x)}{ f (x)+ f ( a + b -x)} d x=\frac{1}{2}(b- a )$ $\therefore \quad \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{1+\sqrt{\cot x}}=\frac{1}{2}\left(\frac{\pi}{3}-\frac{\pi}{6}\right)=\frac{\pi}{12}$
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