When it comes to integrals of trig functions, I always find it helpful to play around with the Pythagorean Identities and see if I get a meaningful result.
For #intsin^3x/cos^2xdx#, we'll make use of the identity #sin^2x+cos^2x=1#, or equivalently, #1-cos^2x=sin^2x#.
Note that we can rewrite the integrand as:
#int((sin^2x)(sinx))/cos^2xdx#
Because #sin^2x=1-cos^2x#, we have:
#int((1-cos^2x)(sinx))/cos^2xdx#
Doing a little algebra,
#=int((1-cos^2x))/cos^2x(sinx)dx#
#=int(1/cos^2x-cos^2x/cos^2x)(sinx)dx#
#=int(1/cos^2x-1)(sinx)dx#
#=intsinx/cos^2x-sinxdx#
Using the sum rule for integrals, this boils down to:
#intsinx/cos^2xdx-intsinxdx#
For the first of these integrals, we can apply a #u#-substitution:
Let #u=cosx#
Then #(du)/dx=-sinx->du=-sinxdx#
Balancing the integral to fit the #u#-substitution, and then applying it, we see:
#intsinx/cos^2xdx#
#=-int(-sinx/cos^2x)dx#
#=-int(du)/u^2#
#=-intu^(-2)du#
#=-(-u^(-1))-># using reverse power rule
#=1/u=1/cosx=secx-># because #u=cosx#
For the other integral, #intsinxdx#, we don't have to do as much work because it's simply #-cosx#.
Putting it all together, we have:
#intsin^3x/cos^2xdx=intsinx/cos^2xdx-intsinxdx=secx-(-cosx)=secx+cosx#
Adding the constant of integration #C#, our final result is:
#intsin^3x/cos^2xdx=secx+cosx+C#
