#int 1/(3+5sinx+3cosx)dx#?

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1 Answer
maganbhai P.
Jun 18, 2018

#I=1/5ln|5tan(x/2)+3|+c#

Explanation:

Here,

#I=int1/(3+5sinx+3cosx)dx#

Subst. #tan(x/2)=t=>sec^2(x/2)*1/2dx=dt#

#=>(1+tan^2(x/2))dx=2dt=>dx=2/(1+t^2)dt#

So,

#I=int1/(3+5((2t)/(1+t^2))+3((1-t^2)/(1+t^2)))xx2/(1+t^2)dt#

#=int2/(3+3t^2+10t++3-3t^2)dt#

#=int2/(10t+6)dt#

#=int1/(5t+3)dt#

#=1/5ln|5t+3|+c#

Subst. back , #t=tan(x/2)#

#I=1/5ln|5tan(x/2)+3|+c#

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