Question

How do you integrate sinθ x secθ cubed?

Answer 1

Integration by parts. let u=sin(theta) dv=sec^3. Find du and v then put into ibp form. After a couple go rounds of doing ibp you will eventually end up with what you started with (sin(theta)sec^3)). Combine all like terms then solve for sin (theta)sec^3.

Answer 2

Thank you. Gosh but it just gets so complicated after one round of ibp that I get lost. That's a nasty integration.

Answer 3

You mean integrate sin(θ)/cos(θ)³ ? can be done directly, think about it, cos(θ)^(-3)*sin(θ), up to a constant, it is u*u'\[\int\limits_{}^{}1\div 3*\cos ^{-3}(\theta)\times3\sin (\theta)\]

Answer 4

sin(x) * sec^3(x)=sin(x)/cos^3(x) trig identity
sin(x)dx is almost the derivative for cosx so as milk_wind suggests you can do a u substitution.
u= cos (x), du = -sin(x)dx or -du = sin(x)dx
so the original problem can be written
integrate -du/u^3 or -u^-3du
when you integrate you get 1/2*u ^-2 or 1/2 cos(x)^-2 or 1/2sec^2(x)

Answer 5

sin(x)/cos^3(x)=tan(x)sec^2(x)
\[\int\limits_{}^{}\tan(x)d(\sec^2(x))\]
if you use u , then "you" not seeing it
since tan(x) derivative is sitting in the integrand you simply raise it to the next power
1/2(tan(x))^2+C