Evaluate (i) `inttan^(3//2) xsec^2x\ dx` (ii) `int(x^3)/((x^2+1)^3)\ dx`

Explanation

Solution
(i) Suppose `I =int tan ^{3//2} x sec ^{2} x d x`
Suppose `tan x=t`, now
`sec ^{2} x d x=d t`
On substituting the value of `x`
`I=int t^{3//2} d t`
On integrating
`I=t^{5//2} / (5//2)+c`
On substituting the value of `t`
`I=2/5(tan x)^(5/2) +c`


` (ii) int(x^3)/((x^2+1)^3)\ dx`
taking `1+x^2=t`
`=>2xdx=dt`
`xdx=(dt)/2`
so, `int (t-1)/(2t^3) dt`
`=1/2 [int t/t^3 dt - int 1/t^3 dt ]`
`=1/2 [[1/(2t^2)-1/t]+c`
`=1/2[1/(2(1+x^2)^2)-1/(1+x^2)]+c`