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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.

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

Answers (1)

Answer:

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

Given:

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

Hint:

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

Explanation:

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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