How do you evaluate the integral of #int (sinx+sinxtan^2x)/sec^2x# to 0 to pi/3?

1 Answer
mason m
Sep 10, 2016

#1/2#

Explanation:

We have:

#I=int_0^(pi/3)(sinx+sinxtan^2x)/sec^2xdx#

#=int_0^(pi/3)(sinx+sinx(sin^2x/cos^2x))/(1/cos^2x)dx#

#=int_0^(pi/3)cos^2x(sinx+sin^3x/cos^2x)dx#

#=int_0^(pi/3)(cos^2xsinx+sin^3x)dx#

Let #sin^3x=sin^2xsinx=(1-cos^2x)sinx=sinx-cos^2xsinx#:

#I=int_0^(pi/3)(cos^2xsinx+sinx-cos^2xsinx)dx#

#=int_0^(pi/3)sinxdx#

Note that #intsinxdx=-cosx+C#:

#I=[-cosx]_0^(pi/3)#

#=-cos(pi/3)-(-cos(0))#

#=-1/2-(-1)#

#=1/2#

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