Evaluate: (i) `intx^3sinx^4\ dx` (ii) `inte^(-x)\ cos e c^2(2\ e^(-x)+5)\ dx` (iii) `int(x^2tan^(-1)x^3)/(1+x^6)\ dx`

Explanation

`(i)` Let `I= intx^3sinx^4\ dx`
Let `x^4=t`
`=>4x^3dx=dt`
`=>dx=1/(4x^3)dt`
therefore
`I= intx^3sintxx(dt)/(4x^3)`
`=1/4 intsin t dt `
`=-1/4cost +c`
`=-1/4 cos x^4 +c`


`(ii)`let `I=inte^(-x)\ cos e c^2(2\ e^(-x)+5)\ dx`
putting `t=2\ e^(-x)+5`
`=>-2e^(-x)dx=dt`
so,`I=-1/2int cosec^2t dt`
`=1/2 cot t+c`
`=1/2 cot (2\ e^(-x)+5)+c`


`(iii)` We have ,
`I=int(x^2tan^(-1)x^3)/(1+x^6)\ dx`
let `t=tan^-1 x^3`
`=>(dt)/(dx)=1/(1+(x)^2)xx3x^2`
`=>(dt)/3=x^2/(1+x^6)`
Therefore,
`I=1/3 int t dt`
`=1/3 t^2/2+c`
On putting the value of t, we get
`I=1/6(tan^-1 x^3)^2+c`