Question Bank [2022] — Maths STD 12 Science — Question
Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]3 Marks
Question
Evaluate: $\int_0^{ a } \frac{1}{x+\sqrt{ a ^2-x^2}} d x$
✓
Answer
Let $I =\int_0^{ a } \frac{1}{x+\sqrt{ a ^2-x^2}} d x$
Put $x=a \sin \theta$
$\therefore dx = a \cos \theta d \theta$
When $x =0, \theta=0$ and when $x = a , \theta=\frac{\pi}{2}$
$ \therefore I =\int_0^{\frac{\pi}{2}} \frac{ a \cos \theta d \theta}{ a \sin \theta+\sqrt{ a ^2- a ^2 \sin ^2 \theta}}$
$=\int_0^{\frac{\pi}{2}} \frac{ a \cos \theta d \theta}{ a \sin \theta+ a \sqrt{1-\sin ^2 \theta}}$
$\int_0^{\frac{\pi}{2}} \frac{\cos \theta}{\sin \theta+\sqrt{\cos ^2 \theta}} d \theta$
$\therefore I =\int_0^{\frac{\pi}{2}} \frac{\cos \theta}{\sin \theta+\cos \theta} d \theta\ldots(i)$
$\therefore I =\int_0^{\frac{\pi}{2}} \frac{\cos \left(\frac{\pi}{2}-\theta\right)}{\sin \left(\frac{\pi}{2}-\theta\right)+\cos \left(\frac{\pi}{2}-\theta\right)}$
${\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]} $
$\therefore I =\int_0^{\frac{\pi}{2}} \frac{\sin \theta}{\cos \theta+\sin \theta} d \theta\ldots(ii)$
Adding (i) and (ii), we get
$ 2 I =\int_0^{\frac{\pi}{2}} \frac{\cos \theta}{\sin \theta+\cos \theta} d \theta+\int_0^{\frac{\pi}{2}} \frac{\sin \theta}{\cos \theta+\sin \theta} d \theta$
$=\int_0^{\frac{\pi}{2}} \frac{\cos \theta+\sin \theta}{\sin \theta+\cos \theta} d \theta$
$=\int_0^{\frac{\pi}{2}} d \theta-[\theta]_0^{\frac{\pi}{2}}$
$=\frac{\pi}{2}-0$
$\therefore I =\frac{1}{2} \times \frac{\pi}{2}$
$\therefore I =\frac{\pi}{4} $
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