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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 23 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 23 Maths Textbook Solution.

Answers (1)

Answer:

-\log |\sin x+\cos x+\sqrt{\sin 2 x}|+c

Given:

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

Hint:

To solve this equation we use sin x formula and substitute method.

Solution: 

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

                                                                        \left[\sin 2 x=(\cos x+\sin x)^{2}-1, \quad t=\cos x+\sin x, \quad d t=(-\sin x+\cos x) d x\right.

=\int \frac{\sin x-\cos x}{\left(\sqrt{ \left.(\cos x+\sin x)^{2}-1\right)}\right)} d x

Let  t=\cos x+\sin x, \quad d t=(-\sin x+\cos x) d x

=-\frac{d t}{\sqrt{t^{2}-1}}                                          \left[\because \frac{1}{\sqrt{t^{2}-a^{2}}} d t=\log \left|t+\sqrt{t^{2}-a^{2}}\right|+c\right.

=-\log \left|t+\sqrt{t^{2}-1}\right|+c

=-\log \left|\sin x+\cos x+\sqrt{(\sin x+\cos x)^{2}-1}\right|+c

=-\log |\sin x+\cos x+\sqrt{\sin 2 x}|+c

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