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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.

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

Answers (1)

Answer:

-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

Given:

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

Hint:

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

Solution: 

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

 

Posted by

infoexpert21

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