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Question
Evaluate : `int1/(3+5cosx)dx`
Solution
Let ` I=int 1/(3+5cosx)dx " put " tan (x/2)=t`
then `dx=2/(1+t^2)dt and cos x=(1-t^2)/(1+t^2)`
`I=int(2dt/(1+t^2))/(3+5((1-t^2)/(1+t^2)))`
`=2int (dt/(1+t^2))/((3(1+t^2)+5(1-t^2))/(1+t^2))`
`=2int dt/(3+3t^2+5-5t^2)`
`=2int dt/(8-2t^2)`
`=int dt/(2^2-t^2) `
`=1/(2(2)) log|(2+t)/(2-t)|+c`
`=1/4 log|(2+tan(x/2))/(2-tan(x/2))|+c`
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