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Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 119 Maths Textbook Solution.

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Answer: -e^{-x / 2} \sec \frac{x}{2}+C

Hint : To solve this equation we have to use differentiation method

Given: \int \frac{\sqrt{1-\sin x}}{1+\cos x} e^{-x / 2} d x


\int e^{-x}\left(f^{\prime}(x)-f(x) d x=e^{-x} f(x)+C\right.

\operatorname{Let} \frac{x}{2}=t

x=2 t

d x=2 d t

I=\int \frac{\sqrt{1-\sin 2 t}}{1+\cos 2 t} \cdot e^{-t} \cdot 2 d t

I=2 \int e^{-t} \cdot \frac{\sqrt{1-\sin 2 t}}{1+\cos 2 t} \cdot d t

\frac{\sqrt{1-\sin 2 t}}{1+\cos 2 t}=\frac{\sqrt{\sin ^{2} t+\cos ^{2} t-2 \sin t \cdot \cos t}}{2 \cos ^{2} t}

I=\frac{\sqrt{(\sin t-\cos t)^{2}}}{2 \cos ^{2} t}=\frac{\sin t-\cos t}{2 \cos ^{2} t}

=\frac{1}{2}\left(\frac{\sin t}{\cos ^{2} t}-\frac{\cos t}{\cos ^{2} t}\right)

=\frac{1}{2}\left(\frac{\tan t}{\cos t}-\frac{1}{\cos t}\right)

=\frac{1}{2}(\tan t \cdot \sec t-\sec t)

f(t)=\frac{1}{2} \sec t ; f^{\prime}(t)=\frac{1}{2} \tan t \sec t

\frac{\sqrt{1-\sin 2 t}}{1+\cos 2 t}=f^{\prime}(t)-f(t)

I=-2 e^{-t} f(t)+C

I=-2 e^{-t} \cdot \frac{1}{2} \sec t+C

I=-e^{-x / 2} \sec \frac{x}{2}+C

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