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Q19   Find the integrals of the functions \frac{1}{\sin x \cos ^3 x }

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best_answer

Let 
We can write 1 = \sin^2x +\cos^2x
Then, the equation can be written as  
I =\frac{\sin^2x +\cos^2x}{\sin x\cos^3x}

I =\int (\tan x+\frac{1}{\tan x})\sec^2 x dx
put the value of tanx = t
So, that \sec^2xdx =dt

\\\Rightarrow I=\int (t+\frac{1}{t})dt\\ =\frac{t^2}{2}+\log\left | t \right |+C\\ =\log\left | \tan x \right |+\frac{1}{2}\tan^2x+C

Posted by

manish

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