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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 49 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 49 Maths Textbook Solution.

Answers (1)

Answer:

I=\ln \left|\sin x+\frac{1}{2}+\sqrt{\sin ^{2} x+\sin x}\right|+c

Given:

\int \sqrt{\operatorname{cosec} x-1} d x

Hint:

To solve this statement we have to convert cosec into sin and then we apply standard formula.

Solution: 

\int \sqrt{\operatorname{cosec} x-1} d x

   I=\int \sqrt{\frac{1}{\sin x}-1} d x

    I=\int \sqrt{\frac{1-\sin x}{\sin x}}-d x

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

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

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

   I=\int \frac{\cos x}{\sqrt{\sin x(1+\sin x)}} d x

 sin x = t ,

dt = cos x dx

\therefore I=\int \frac{d t}{\sqrt{t}(t+1)}=\int \frac{d t}{\sqrt{t^{2}+t}}=\int \frac{d t}{\sqrt{\left(t+\frac{1}{2}\right)^{2}-\frac{1}{4}}}=\int \frac{d t}{\sqrt{\left(t+\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}}

I=\ln \left|t+\frac{1}{2}+\sqrt{t^{2}+t}\right|+c \quad\left[\because \int \frac{d x}{\sqrt{x^{2}+a^{2}}}=\ln \left|x+\sqrt{x^{2}+a^{2}}\right|+c\right]

I=\ln \left|\sin x+\frac{1}{2}+\sqrt{\sin ^{2} x+\sin x}\right|+c

 

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