Evaluate :
(i) `int(1)/((1+tanx))dx` (ii) `int(1)/((1+cotx))dx`
(iii) `int((1-tanx)/(1+tanx))dx` (iv) `int(tanx)/((secx+cosx))dx`

Explanation

(i) `int(1)/((1+tanx))dx=int(1)/((1+(sinx)/(cosx)))dx`
`=int(cosx)/((cosx+sinx))dx=int((cosx+sinx)+(cosx-sinx))/(2(cosx+sinx))dx`
`=(1)/(2)intdx+(1)/(2)int((cosx-sinx))/((cosx+sinx))dx`
`=(1)/(2)intdx+(1)/(2)int(1)/(t)dt," where"(cosx+sinx)=tand(cosx-sinx)dx=dt`
`=(1)/(2)x+(1)/(2)log|t|+C=(1)/(2)x+(1)/(2)log|{:cosx+sinx:}|+C`
(ii) `int(1)/((1+cotx))dx=int(1)/((1+(cosx)/(sinx)))dx=int(sinx)/((sinx+cosx))dx`
`=int((sinx+cosx)-(cosx-sinx))/(2(sinx+cosx))dx`
`=(1)/(2)intdx-(1)/(2)int((cosx-sinx))/((sinx+cosx))dx`
`=(1)/(2)intdx-(1)/(2)int(1)/(t)dt`,
where sin x + cos x = t and (cos x - sin x)dx=dt
`=(1)/(2)x-(1)/(2)log|t|+C=(1)/(2)x-(1)/(2)log|{:sinx+cosx:}|+C`.
(iii) `int((1-tanx)/(1+tanx))dx=int((1-(sinx)/(cosx)))/((1+(sinx)/(cosx)))dx=int((cosx-sinx))/((cosx+sinx))dx`
`=int(1)/(t)dt," where"(cosx+sinx)=t and(cosx+sinx)dx=dt`
`=log|t|+C=log|{:(cosx+sinx):}|+C`.
(iv) `int(tanx)/((secx+cosx))dx=int(sinx)/(1+cos^(2)x)dx`
`=-int(1)/((1+t^(2)))dt," where"cosx=t and sinx dx=-dt`,
`=-tan^(-1)t+C=-tan^(-1)(cosx)+C`.