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# find integral :- root cos(2x)/sin x Grade:11

## 1 Answers

10 years ago

put cos2x=1-tan^2x /1+tan^2x

expression becomes root(1-tan^2x)/tanx

now multiply and divide by 1+tan^2x

now  expression  is root(1-tan^2x) .sec^2x /(1+tan^2x)tanx

now put tanx=t  , we get sec^2xdx=dt

now expression is root(1-t^2)/t.(1+t^2)

now multiply divide by t

expression is root (1-t^2)t/t^2 .(1+t^2)

now put t^2=u, so 2tdt=du

now expression is root1-u/2u(1+u)

now again put 1-u=v^2 ,so -du=2vdv

expression is -v^2/(1-v^2)(2-v^2)

=1/2-v^2  +  -1/(2-v^2)(1-v^2)

now integrate using partial fraction and get the answer

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