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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.

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


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


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


\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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