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Find the integral $\large \int \frac{\sec^2x}{\large cosec^2x}$$dx$

This question has appeared in model paper 2012.

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  • $\sec x=\large \frac{1}{\cos x} $ and cosec $x=\large \frac{1}{\sin x}.$
$\large\int \frac{\sec^2x}{cosec^2x} $$dx.$
We know that $\sec x=\large \frac{1}{\cos x} $ and cosec $x=\large \frac{1}{\sin x}.$
$\rightarrow \large\int \frac{\sec^2x}{cos^2x}$$dx=\large\int $$ \tan^2x \;dx.$
But since $\tan^2x=(\sec^2x-1)$.
$\large\int $$ \tan^2x \;dx.=\large \int$$(\sec^2x-1)dx =\large \int $$\sec^2xdx-$$\large \int$$ dx.$
$\Rightarrow \large\int $$ \tan^2x \;dx.$$=\tan x-x+c.$
$\Rightarrow \large \int\frac{\sec^2x}{cosec^2x}$$dx=\tan x-x+c.$
answered Dec 20, 2013 by balaji.thirumalai
 
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