Evaluate : `int(3cosx+2)/(sinx+2cosx+3)` dx

Explanation

We have,
`I=int(3cosx+2)/(sinx+2cosx+3)`dx
Let `3cosx+2=lambda(sinx+2cosx+3) + mu(cosx-2sinx)+v`
Comparing the coefficients of sinx, cosx and constant term on both sides, we get
`lambda-2mu=0, 2lambda+mu=3, 3lambda+v=2 rArr lambda=6/5, mu=3/5` and `v=-8/5`
`therefore i=int(lambda(sinx+2cosx+3)+mu(cosx-2sinx)+v)/(sinx+2cosx+3)`dx ltrgt `rArr I=lambda intdx + mu int(cosx-2sinx)/(sinx+2cosx+3)dx+v int1/(sinx+2cosx+3)dx`
`rArr I=lambda+mulog|sinx+2cosx+3|+vI_(1)`
Where `I_(1)=int1/(sinx+2cosx+3)dx`
Putting, `sinx=(2tanx//2)/(1+tan^(2)x//2), cosx=(1-tan^(2)x//2)/(1+tan^(2)x//2)`, we get
`I_(1)=int1/((2tanx//2)/(1+tan^(2)x//2)+(2(1-tan^(2)x//2))/(1+tan^(2)x//2)+3` dx `=int(1+tan^(2)x//2)/(2tanx//2+2-2tan^(2)x/2+3(1+tan^(2)x//2))`dx
`int(sec^(2)x//2)/(tan^(2)x//2+2tanx//2+5)`dx
Putting `tanx/2=t` and `1/2sec^(2)x/2=dt` or `sec^(2)x/2dx=2dt`, we get
`I_(1)= int(2dt)/(t^(2)+2t+5)= 2int(dt)/((t+1)^(2)+2^(2))=2/2tan^(-1)(t+1)/(2)= tan^(-1)(tanx/2+1)/(2)`
Hence, `I=lambdax + mulog |sinx+2cosx+3|+v tan^(-1)(tanx/2+1)/(2)+C`
Where `lambda=6/5, mu=3/5` and `v=-8/5`