Evaluate: (i) `int(sinx)/((1+cosx)^2)\ dx` (ii) `intcotxlogsinx\ dx`

Explanation

(i) `I=int(sinx)/((1+cosx)^2)\ dx`
Let `t=1+cosx`
`=-sinx dx=dt`
so,`I= int(-dt)/t^2`
`=-1/t +c`
`=-1/(1+cosx)+c`

(ii) `I=intcotxlogsinx\ dx`
Let `t= log sinx`
`=>1/sinx cosx dx =dt`
`=>cot x dx =dt`
so,`I= int tdt`
`=t^2/2+c`
`=|log sinx |^2 /2+c`