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Provide Solution For  R. D. Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Multiple Choice Questions Question 28 Maths Textbook Solution.

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\pm \log (\sin x-\cos x)+C


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


You must know about the derivation of sin and cos function and \int \frac{1}{x} d x .


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

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

          =\int \frac{\sin x+\cos x}{\sqrt{(\cos x-\sin x)^{2}}} d x                                                        \left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]

          \begin{aligned} &=\int \frac{\sin x+\cos x}{(\cos x-\sin x)} d x \\ &=\int \frac{d t}{ t} \end{aligned}                                                                      [\text { Put } \cos x-\sin x=t \Rightarrow(\cos x+\sin x) d x=d t]

          \begin{aligned} &= \log |t|+C \\ &= \log (\cos x-\sin x)+C \end{aligned}

Hence,\int \frac{\sin x+\cos x}{\sqrt{1-\sin 2 x}}= \log (\cos x-\sin x)+C .

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