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Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 26 Maths Textbook Solution.

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-2 \sqrt{1-\sin x}+\sqrt{2} \ln \left|\sec \left(\frac{\pi}{4}-\frac{x}{2}\right)+\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right|+c


\int \frac{\sin x}{\sqrt{1+\sin x}} d x


To solve the statement we have to use formula like sin x and 1+cos x


I=\int \sqrt{\sin x+1}-\frac{1}{\sqrt{\sin x+1}} d x

   =\int \sqrt{\sin x+1} d x-\int \frac{d x}{\sqrt{\sin x+1}}

   I_{1}=\int \sqrt{\sin x+1} d x

      =\int \sqrt{1+\sin x} \cdot \frac{\sqrt{1-\sin x}}{\sqrt{1-\sin x}} d x

   =\frac{\sqrt{(1+\sin x)(1-\sin x)}}{\sqrt{1-\sin x}} d x

  =\frac{\sqrt{\cos ^{2} x}}{\sqrt{1-\sin x}} d x

 =\frac{\cos x}{\sqrt{1-\sin x}} d x

 =\sqrt{1-\sin x}=u

\frac{1}{2 \sqrt{1-\sin x}} \cos x d x=d u

\frac{\cos x}{\sqrt{1-\sin x}}=-2 d u

I_{1}=\int-2 d u=-2 u=-2 \sqrt{1-\sin x}

I_{2}=\int \frac{d x}{\sqrt{1+\sin x}}=\int \frac{d x}{\sqrt{1+\cos \left(\frac{\pi}{2}-x\right)}}                            \left[\because \sin x=\cos \left(\frac{\pi}{2}-x\right)\right],\left[1+\cos x=2 \cos ^{2} x \frac{x}{2}\right]

=\int \frac{d x}{\sqrt{2 \cos ^{2}\left(\frac{\pi}{4}-\frac{x}{2}\right)}}

=\int \frac{d x}{\sqrt{2}\left(\cos \left(\frac{\pi}{4}-\frac{x}{2}\right)\right.}                                            \left[\because \int \sec x d x=\ln |\sec x+\tan x|+c\right]

=\int \frac{1}{\sqrt{2}} \sec \left(\frac{\pi}{4}-\frac{x}{2}\right) d x

=\frac{1}{\sqrt{2}} \cdot \frac{1}{-\frac{1}{2}} \ln \left|\sec \left(\frac{\pi}{4}-\frac{x}{2}\right)+\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right|+c


=-\sqrt{2} \ln \left|\sec \left(\frac{\pi}{4}-\frac{x}{2}\right)+\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right|+c

\mathrm{I}_{1}-\mathrm{I}_{2}=\mathrm{l}=-2 \sqrt{1-\sin x}+\sqrt{2} \ln \left|\sec \left(\frac{\pi}{4}-\frac{x}{2}\right)+\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right|+c


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