Question

How do I find `inttan^4x dx`?

Answer 1

Well, this one is quite difficult;

`inttan^4(x)dx=int tan^2(x)tan^2(x)dx=`
`=intsin^2(x)/cos^2(x)tan^2(x)dx=`
`=int(1-cos^2(x))/cos^2(x)tan^2(x)dx=`
`=int[1/cos^2(x)-1]tan^2(x)dx=`
`=int[tan^2(x)/cos^2(x)-tan^2(x)]dx=`
But `d[tan(x)]=1/cos^2(x)dx`

`=inttan^2(x)d[tan(x)]-inttan^2(x)dx=`
`=tan^3(x)/3-inttan^2(x)dx=`
`=tan^3(x)/3-intsin^2(x)/cos^2(x)dx=`
`=tan^3(x)/3-int[(1-cos^2(x))/cos^2(x)]dx=`
`=tan^3(x)/3-{int1/cos^2(x)dx-intdx}=`
`=tan^3(x)/3-tan(x)-x+c`