How do I evaluate inttan^3(x) sec^5(x)dx∫tan3(x)sec5(x)dx?

1 Answer
Massimiliano
Feb 4, 2015

The answer is: 1/7cos^-7x-1/5cos^-5x+c17cos−7x−15cos−5x+c

We can write this integral in this way:

int(sin^3x)/cos^3x*1/cos^5xdx=intsin^3x/cos^8xdx=∫sin3xcos3x⋅1cos5xdx=∫sin3xcos8xdx=

=int sin^2xsinxcos^-8xdx=int(1-cos^2x)sinxcos^-8xdx==∫sin2xsinxcos−8xdx=∫(1−cos2x)sinxcos−8xdx=

=intsinxcos^-8xdx-intsinxcos^-6xdx==∫sinxcos−8xdx−∫sinxcos−6xdx=

=-int-sinxcos^-8xdx+intsinxcos^-6xdx==−∫−sinxcos−8xdx+∫sinxcos−6xdx=

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

=1/7cos^-7x-1/5cos^-5x+c=17cos−7x−15cos−5x+c.

I have used the integral:

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

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