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aditya prakash morey Grade:
        

find integral :- root cos(2x)/sin x 

6 years ago

Answers : (1)

vikas askiitian expert
510 Points
										

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

6 years ago
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