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

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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#.