How do I evaluate $inttan^3(x) sec^5(x)dx$ ?

Question

1818 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-02-04T18:48:55

The answer is: $1/7cos^-7x-1/5cos^-5x+c$

We can write this integral in this way:

$int(sin^3x)/cos^3x*1/cos^5xdx=intsin^3x/cos^8xdx=$

$=int sin^2xsinxcos^-8xdx=int(1-cos^2x)sinxcos^-8xdx=$

$=intsinxcos^-8xdx-intsinxcos^-6xdx=$

$=-int-sinxcos^-8xdx+intsinxcos^-6xdx=$

$=-cos^(-8+1)x/(-8+1)+cos^(-6+1)x/(-6+1)+c=-cos^-7x/(-7)+cos^-5x/-5+c=$

$=1/7cos^-7x-1/5cos^-5x+c$ .

I have used the integral:

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

How do I evaluate inttan^3(x) sec^5(x)dxinttan^3(x) sec^5(x)dx?