Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Integrate: 1/4sin^2x+9cos^2x

Answers (1)

best_answer

Solution:

                \Rightarrow I=int dx/4sin^2x+9cos^2x

        Here dividing N^r and D^r by cos^2x . We get,

                \Rightarrow I=int sec^2xhspace0.1cmdx/4tan^2x+9

            Put ,          tanx=tRightarrow sec^2xhspace0.1cmdx=dt

 

          \I=int dt/4t^2+9=1/4int dt/t^2+(3/2)^2\ \Rightarrow I=1/6 an^-1(2tanx/3)+c 

Posted by

Deependra Verma

View full answer

Similar Questions

Latest Question