Evaluate : `(i) int_(1)^(3)(cos(logx))/(x)dx` `(ii) int_(0)^(pi//2)sqrt(cos theta)sin^(3)theta d theta` `(iii) int_(0)^(pi//2)(cosx)/((1+sinx)(2+sinx))dx`

Explanation

`(i)` Put `logx=t` so that `(1)/(x)dx=dt`.
Also, `(x=1impliest=log1=0)` and `(x=3impliest=log3)`.
`:. Int_(1)^(3)(cos(logx))/(x)dx=int_(0)^(log3)costdt=[sint]_(0)^(log3)=sin(log3)`.
`(ii)` Put `cos theta=t` so that `sin theta d theta=-dt`.
Also, `(theta=0impliest=1)` and `(theta=(pi)/(2)impliest=0)`
`:.int_(0)^(pi//2)sqrt(costheta)sin^(3)d theta=int_(0)^(pi//2)sqrt(costheta)*(1-cos^(2)theta)sin theta d theta`
`=-int_(1)^(0)sqrt(t)(1-t^(2))dt=int_(0)^(1)(t^(1//2)-t^(5//2))dt`
`=[(2)/(3)t^(3//2)-(2)/(7)t^(7//2)]_(0)^(1)=((2)/(3)-(2)/(7))=(8)/(21)`.
`(iii)` Put `sinx=t` so that `cosx dx=dt`.
Also, `(x=0impliest=0)` and `(x=(pi)/(2)impliest=1)`.
`:.int_(0)^(pi//2)(cosx)/((1+sinx)(2+sinx))dx`
`=int_(0)^(1)(dt)/((1+t)(2+t))`
`=int_(0)^(1)[(1)/((1+t))-(1)/((2+t))]dt` [ by partial fractions]
`=int_(0)^(1)(dt)/((1+t))-int_(0)^(1)(dt)/((2+t))`
`=[log|1+t|]_(0)^(1)-[log|2+t|]_(0)^(1)`
`=[(log2-log1)-(log3-log2)]=(2log2)-(log3)`.
`(iv) int_(0)^(pi//2)(dx)/((1-2sinx))=int_(0)^(pi//2)(dx)/(1-2{(2tan(x//2))/(1+tan^(2)(x//2))})`
`=int_(0)^(pi//2)(sec^(2)(x//2))/([1+tan^(2)(x//2)-4tan(x//2)])dx`
`=2int_(0)^(1)(dt)/((1+t^(2)-4t))`, where `tan.(x)/(2)=t` `[{:(x=0impliest=0),(x=(pi)/(2)impliest=1):}]`
`=2int_(0)^(1)(dt)/((t-2)^(2)-(sqrt(3))^(2))=2*(1)/(2sqrt(3))[log|(t-2sqrt(3))/(t-2+sqrt(3))|]_(0)^(1)`
`=(1)/(sqrt(3))[log((sqrt(3)+1)/(sqrt(3)-1))-log((sqrt(3)+2)/(sqrt(3)-2))]`.
`(v) int_(0)^(pi//2)(dx)/((3+2cosx))=int_(0)^(pi//2)(dx)/(3+2*[(1-tan^(2)(x//2))/(1+tan^(2)(x//2))])`
`=int_(0)^(pi//2)(sec^(2)(x//2))/(tan^(2)(x//2)+5)dx`
`=2int_(0)^(1)(dt)/(t^(2)+(sqrt(5))^(2))`, where `tan.(x)/(2)=t` `[{:(x=0impliest=0),(x=(pi)/(2)impliest=1):}]`
`=2*(1)/(sqrt(5))[tan^(-1).(t)/(sqrt(5))]_(0)^(1)=(2)/(sqrt(5))tan^(-1).(1)/(sqrt(5))`.
`(vi) int_(0)^(pi//2)(dx)/((4sin^(2)x+5cos^(2)x))=int_(0)^(pi//2)(sec^(2)x)/((4tan^(2)x+5))dx`. [dividing num. and denom. by `cos^(2)x`]
`=int_(0)^(oo)(dt)/((4t^(2)+5))`, where `tanx=t` `[{:(x=0impliest=0),(x=(pi)/(2)impliest=oo):}]`
`=(1)/(4)int_(0)^(oo)(dt)/(t^(2)+((sqrt(5))/(2))^(2))=(1)/(4)*(2)/(sqrt(5))[tan^(-1).(2)/(sqrt(5))]_(0)^(oo)`
`=(1)/(2sqrt(5))[tan^(-1)(oo)-tan^(-1)(0)]`
`=(1)/(2sqrt(5))((pi)/(2)-0)=(pi)/(4sqrt(5))`.