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  • Explain Solution R.D. Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 83 Maths Textbook Solution.

Explain Solution R.D. Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 83 Maths Textbook Solution.

Answers (1)

Answer:

-\frac{1}{2} \cos x+\frac{1}{2} \sin x-\frac{1}{2 \sqrt{2}} \log \left(\operatorname{cosec}\left(x+\frac{\pi}{4}\right)-\cot \left(x+\frac{\pi}{4}\right)\right)+c

Hint:

You must know about integration of sec x & cosec x.

Given:

\int \frac{1}{\sec x+\operatorname{cosec} x} d x

Soltuion:

\operatorname{now}, \int \frac{1}{\frac{1}{\cos x}+\frac{1}{\sin x}} d x \quad\left[\because \sec x=\frac{1}{\cos x}, \operatorname{cosec} x=\frac{1}{\sin x}\right]

=\int \frac{d x(\cos x \sin x)}{\sin x+\cos x}

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

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

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

=\frac{1}{2} \int(\sin x+\cos x)=d x-\int \frac{1}{\sin x+\cos x} d x

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

=\frac{1}{2}\left[-\cos x+\sin x-\frac{1}{\sqrt{2}} \int \frac{d x}{\sin \left(x+\left(\frac{\pi}{4}\right)\right.}\right]

=\frac{1}{2}\left[-\cos x+\sin x-\frac{1}{\sqrt{2}} \int \operatorname{cosec}\left(x+\frac{\pi}{4}\right) d x\right]

=-\frac{1}{2} \cos x+\frac{1}{2} \sin x-\frac{1}{2 \sqrt{2}} \log \left(\operatorname{cosec}\left(x+\frac{\pi}{4}\right)-\cot \left(x+\frac{\pi}{4}\right)\right)

=-\frac{1}{2} \cos x+\frac{1}{2} \sin x-\frac{1}{2 \sqrt{2}} \log \left(\operatorname{cosec}\left(x+\frac{\pi}{4}\right)-\cot \left(x+\frac{\pi}{4}\right)\right)+c

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