If `int1/(5+4sinx)\ dx=Atan^(-1)(Btanx/2+4/3)+C` , then `A=2/3,\ \ B=5/3` (b) `A=1/3,\ \ B=2/3` (c) `A=-2/3,\ \ B=5/3` (d) `A=1/3,\ \ B=-5/3`

Explanation

Given, `int1/(5+4sinx)dx`

= `int1/(5+4(2tan(x/2))/(1+tan^2(x/2))) dx`

=`(1+tan^2(x/2))/(5+5tan^
(x/2)+8tan(x/2))dx`

Let `t=tan(x/2)​⇒dt=1/2sec^2(x/2)​dx`

=`∫(2dt​)/(5t^2+8t+5)`

=`2∫dt/(5(t^2+8/5​t+1)`

=`2/5​∫dt/(t^2+2×4/5​t+16/25​−16/25​+1)dt​​`

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