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Provide Solution For  R.D. Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.18 Question 18 Maths Textbook Solution.

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Answer:\sin ^{-1}\left[\frac{1}{3}(\sin x+\cos x)\right]+c

Hint  \int \frac{d t}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1} \frac{x}{a}

Given: \int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x

Explanation:

      \int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x

    =\int \frac{\cos x-\sin x}{\sqrt{8+1-1-\sin 2 x}} d x

    =\int \frac{\cos x-\sin x}{\sqrt{9-\left(\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x\right)}} d x \quad\left[\begin{array}{l} \sin ^{2} x+\cos ^{2} x=1 \\ \sin 2 x=2 \sin x \cdot \cos x \end{array}\right]

    =\int \frac{\cos x-\sin x}{\sqrt{9-(\sin x+\cos x)^{2}}} d x

            Let \cos x+\sin x=t

            (-\sin x+\cos x) d x=d t

Put in (1)

        \int \frac{d t}{\sqrt{3^{2}-t^{2}}}=\sin ^{-1}\left[\frac{t}{3}\right]+c                                                    \left[\int \frac{d t}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1} \frac{x}{a}\right]

        =\sin ^{-1}\left[\frac{1}{3}(\sin x+\cos x)\right]+c

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