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
