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Integration of 1/sinx+cosecx= ????

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Integration of 1/sinx+cosecx= ????

Posted by Arpan Sahu 8 years, 10 months ago

CBSE > Class 12 > Mathematics

1 answers

Naveen Sharma 8 years, 10 months ago

Ans. $\int \frac{1}{\sin x + \cos e c x} d x$

=> $\int \frac{1}{\sin x + \frac{1}{\sin x}} d x$

=> $\int \frac{1}{\frac{\sin^{2} x + 1}{\sin x}} d x$

=> $\int \frac{\sin x}{1 + \sin^{2} x} d x$

=> $\int \frac{\sin x}{2 - \cos^{2} x} d x$

=> $\int \frac{- \sin x}{\cos^{2} x - 2} d x$

Put cos x = t then -sinx dx = dt

=> $\int \frac{1}{t^{2} - 2} d t$

=> $\int \frac{1}{t^{2} - {(\sqrt{2})}^{2}} d t$

We know that : $\int \frac{1}{x^{2} - a^{2}} d x = \frac{1}{2 a} \ln (\frac{(x - a)}{(x + a)}) + C$ , applying this on above. we get

=> $\frac{1}{2 \sqrt{2}} \ln (\frac{t - \sqrt{2}}{t + \sqrt{2}}) + C$

put value of t, we get

=> $\frac{1}{2 \sqrt{2}} \ln (\frac{\cos x - \sqrt{2}}{c o x s + \sqrt{2}}) + C$

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