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Model Paper 10 — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsModel Paper 103 Marks
Question
Evaluate the integral: $\sec ^{-1} \sqrt{x} d x$
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Answer

Let the given integral be,
$I=\int \sec ^{-1} \sqrt{x} d x$
$\text { Putting } \sqrt{x}=\sec \theta$
$\Rightarrow x=\sec ^2 \theta$
$\Rightarrow dx=2 \sec \theta \sec \theta \tan \theta d \theta$
$=2 \sec ^2 \theta \tan \theta d \theta$
$\therefore I=2 \int \theta \sec ^2 \theta \tan \theta d \theta$
$=2 \int \theta \tan \theta \sec ^2 \theta d \theta$
Considering $\theta$ as first function and $\tan \theta \sec ^2 \theta$ as second function
$I=2\left[\theta \frac{\tan ^2 \theta}{2}-\int 1 \frac{\tan ^2 \theta}{2} d \theta\right]\left(\because \int \tan \theta \sec ^2 \tan \theta d \theta=\frac{\tan ^2 \theta}{2}\right)$
$=\theta \tan ^2 \theta-\int\left(\sec ^2 \theta-1\right) d \theta$
$=\theta \tan ^2 \theta-\tan \theta+\theta+C$
$=\theta\left(1+\tan ^2 \theta\right)-\tan \theta+C$
$=\theta \sec ^2 \theta-\sqrt{\sec ^2 \theta-1}+C$
$=\sec ^{-1} \sqrt{x} x-\sqrt{x-1}+C$
$=x \sec ^{-1} \sqrt{x}-\sqrt{x-1}+C$

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