Evaluate : `(i) int_(0)^(2)e^(x//2)dx` `(ii) int_(2)^(4)(x)/((x^(2)+1))dx` `(iii) int_(0)^(1)cos^(-1)xdx`

Explanation

`(i)` Put `(x)/(2)=t` so that `dx=2dt`.
Also, `(x=0impliest=0)` and `(x=2=t=1)`.
`:.int_(0)^(2)e^(x//2)dx=2int_(0)^(1)e^(t)dt=2[e^(t)]_(0)^(1)=2(e-1)`.
`(ii)` Put `(x^(2)+1)=t` so that `xdx=(1)/(2)dt`.
Also, `(x=2impliest=5)` and `(x=4impliest=17)`.
`:.int_(2)^(4)(x)/((x^(2)+1))dx=(1)/(2)int_(5)^(17)(dt)/(t)=(1)/(2)[log|t|]_(5)^(17)=(1)/(2)(log17-log5)`.
`(iii)` Put `x=cost` so that `dx=-sint dt`.
Also, `(x=0impliest=(pi)/(2))` and `(x=1impliest=0)`.
`:.int_(0)^(1)cos^(-1)xdx=-int_(pi//2)^(0)cos^(-1)(cost)sintdt=int_(0)^(pi//2)tsintdt`
`=[t(-cost)]_(0)^(pi//2)-int_(0)^(pi//2)1*(-cost)dt` [integrating by parts]
`=[sint]_(0)^(pi//2)=1`.
`(iv)` Let `(2x+3)-=A*(d)/(dx)(5x^(2)+1)+B`.
Then, `(2x+3)-=(10x)A+B`.
Comparing the coefficients of like powers of `x`, we get
`10A=2` or `A=(1)/(5)` and `B=3`.
`:.(2x+3)=(1)/(5)(10x)+3`.
So, `int_(0)^(1)((2x+3))/((5x^(2)+1))dx=int_(0)^(1)((1)/(5)(10x)+3)/((5x^(2)+1))dx`
`=(1)/(5)(10x)/((5x^(2)+1))dx+3int_(0)^(1)(dx)/((5x^(2)+1))`
`=(1)/(5)[log|5x^(2)+1|]_(0)^(1)+(3)/(5)int_(0)^(1)(dx)/(x^(2)+((1)/(sqrt(5)))^(2))`
`=(1)/(5)log6+(3)/(5)*sqrt(5)[tan^(-1)(x)/((1//sqrt(5)))]_(0)^(1)`
`=(1)/(5)log6+(3)/(sqrt(5))(tan^(-1)sqrt(5))`.