Question

What is the antiderivative of `sec^5xtan^7x`?

Answer 1

The answer is: `sec^11x/11-sec^9x/3+(3sec^7x)/7-sec^5x/5+c`.

Remembering that:

• `sec^2x=1/cos^2x=(sin^2x+cos^2x)/cos^2x=`

`=sin^2x/cos^2x+cos^2x/cos^2x=tan^2x+1rArr`

`tan^2x=sec^2x-1`;

• `intsecxtanxdx=secx+c`;

• `int[f(x)]^n*f'(x)dx=[f(x)]^(n+1)/(n+1)+c`.

Than:

`intsec^5x*tan^7xdx=intsec^5x*tan^6x*tanxdx=`

`=intsec^5x*(sec^2x-1)^3*tanxdx=`

`=intsec^5x(sec^6x-3sec^4x+3sec^2x-1)*tanxdx=`

`=int(sec^11xtanx-3sec^9xtanx+3sec^7xtanx-sec^5xtanx)dx=`

`=intsec^10xsecxtanxdx-3intsec^8xsecxtanxdx+`

`+3intsec^6xsecxtanxdx-intsec^4xsecxtanxdx=`

`=sec^11x/11-(3sec^9x)/9+(3sec^7x)/7-sec^5x/5+c=`

`=sec^11x/11-sec^9x/3+(3sec^7x)/7-sec^5x/5+c`.