Integrating Trigonometric Functions with Powers

[Kuma]

Homework Statement

I want to find the integral from 0 to 2pi of

sin^3 t + t + cos^2 t

Homework Equations

The Attempt at a Solution

I could convert sin^3 t to sin t (sin^2 t) and then use the identity 1 - cos 2t/2 and same for cos^2 t but I was wondering if there is a less messy way to evaluate this.

 

[Dick]

Since your limits are 0 to 2pi if you clever you should be able to come up with an argument about the value of the sin(t)^3 part without integrating it. But you could also use another even simpler identity for sin(t)^2.

 

[Kuma]

Ah, I see that it is 0 for sin^3 t, does this hold for any period of 2pi for sin^3? Also what is the other identity that I could use? I thought it would just be the half angle identity.

 

[Curious3141]

Homework Statement

I want to find the integral from 0 to 2pi of

sin^3 t + t + cos^2 t

Homework Equations

The Attempt at a Solution

I could convert sin^3 t to sin t (sin^2 t) and then use the identity 1 - cos 2t/2 and same for cos^2 t but I was wondering if there is a less messy way to evaluate this.

Integrating the t term is trivial.

The cos^2 term can be handled with the double angle formula for cosine.

The sin^3 term can be immediately reduced to zero by sketching the curve for y = sin^3x and observing that it's an odd periodic function. It applies for any *even* multiple of pi at both upper and lower bounds.

 

[Dick]

Ah, I see that it is 0 for sin^3 t, does this hold for any period of 2pi for sin^3? Also what is the other identity that I could use? I thought it would just be the half angle identity.

sin(t)^2=1-cos(t)^2.