વિધેયનું સંકલન કરો: $\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$

ધારો કે $I = \int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} dx$.
$x = \cos^2 \theta$ લેતા,તેથી $dx = -2 \sin \theta \cos \theta d\theta$.
$I = \int \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} (-2 \sin \theta \cos \theta) d\theta$
$= -2 \int \sqrt{\frac{2 \sin^2 (\theta/2)}{2 \cos^2 (\theta/2)}} \sin \theta \cos \theta d\theta$
$= -2 \int \tan(\theta/2) \cdot (2 \sin(\theta/2) \cos(\theta/2)) \cos \theta d\theta$
$= -4 \int \sin^2(\theta/2) \cos \theta d\theta$
$= -4 \int \left(\frac{1-\cos \theta}{2}\right) \cos \theta d\theta$
$= -2 \int (\cos \theta - \cos^2 \theta) d\theta$
$= -2 \int \cos \theta d\theta + 2 \int \left(\frac{1+\cos 2\theta}{2}\right) d\theta$
$= -2 \sin \theta + \theta + \frac{\sin 2\theta}{2} + C$
$= -2 \sin \theta + \theta + \sin \theta \cos \theta + C$
અહીં $x = \cos^2 \theta$ હોવાથી,$\cos \theta = \sqrt{x}$ અને $\sin \theta = \sqrt{1-x}$ થાય.
$I = -2 \sqrt{1-x} + \cos^{-1}(\sqrt{x}) + \sqrt{1-x} \cdot \sqrt{x} + C$
$= \cos^{-1}(\sqrt{x}) - \sqrt{1-x} (2 - \sqrt{x}) + C$