Question

What is `int xtan^3 (5x)* sec^6 (5x)dx`?

Answer 1

`int x* tan^3 (5x)* sec^6 (5x) dx`

`=x/40 tan^8(5x)-1/1400 tan^7(5x)+x/15 tan^6(5x)-1/600 tan^5(5x)+x/20 tan^4 (5x)-1/1800 tan^3(5x)+1/600 tan(5x)-x/120+C`

Explanation:

Use integration by parts formula `int u dv=uv-int v du`

from the given `int x* tan^3 (5x)* sec^6 (5x) dx`

Let `u = x`
`dv = tan^3(5x)` `sec^6(5x) dx`

expand using Trigonometric identities

`dv=tan^3(5x)` `sec^4(5x)` `sec^2(5x) dx`
`dv=tan^3(5x)(tan^2(5x)+1)^2* sec^2(5x) dx`
`dv= tan^3(5x)[tan^4(5x)+2 tan^2(5x) +1] sec^2(5x) dx`
`dv=[tan^7(5x)+2 tan^5(5x) + tan ^3(5x)] sec^2(5x) dx`

after integration

`v=1/40 tan^8(5x) +1/15 tan ^6(5x) +1/20 tan^4(5x)`

`du = dx`

Use now the formula `int u dv=uv-int v du`