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Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.3 Question 9 Maths Textbook Solution.

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Answer: \frac{-1}{2 \sqrt{2}} \cos 2 x+c

Hint: \text { To solve this equation we use } 1+\cos 2 x \text { and } \sin 2 x \text { formula }

Given: \int \sin x \sqrt{1+\cos 2 x} d x

Solution: \int \sin x \sqrt{1+\cos 2 x} d x

\begin{aligned} &\because 1+\cos 2 x=2 \cos ^{2} x \\ &I=\int \sin x \sqrt{2 \cos ^{2} x} d x \\ &I=\sqrt{2} \int \sin x \cos x d x \\ &I=\frac{\sqrt{2}}{2} \int 2 \sin x \cos x d x \end{aligned}

\begin{aligned} &\because \sin 2 x=2 \sin x \cos x \\ &I=\frac{\sqrt{2}}{2} \int \sin 2 x d x \\ &=\frac{1}{\sqrt{2}}\left[\frac{-\cos 2 x}{2}\right]+c \\ &=\frac{-1}{2 \sqrt{2}} \cos 2 x+c \end{aligned}

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