# How to Integrate sin(x)^3 and cos(x)^3, Sine Cubed and Cosine Cubed

The antiderivatives of sin(x)^3 and cos(x)^3 [also written sin^3 (x) and cos^3 (x)] can be found using a simple trigonometric identity and u-substitution. Some alternative forms of sine cubed and cosine cubed can be found by applying the triple angle formula, which provides another method of integration that some calculus students find easier. The integrals of sin(x)^3 and cos(x)^3 are worked out below, step by step, with the alternative method discussed at the end. These methods can also be extended to work out the antiderivatives of higher odd powers of sine and cosine, such as sin(x)^5, sin(x)^7, etc.

## Integrating sin(x)^3

To find the antiderivative of sin cubed, we note that it can be rewritten as

sin(x) * sin(x)^2

Using the trig formula sin(x)^2 = 1 - cos(x)^2, we can further transform the original function into

sin(x) * [ 1 - cos(x)^2 ]

= sin(x) - sin(x)*cos(x)^2

Though this may look more complicated than what we started with, it is actually much easier to integrate the function in this form. The first part of the function, sin(x), can be integrated as

∫ sin(x) dx = -cos(x)

The second part of the function, -sin(x)*cos(x)^2, lends itself to the calculus technique of u-substitution. Let u = cos(x) and du = -sin(x), then the integral

∫ -sin(x)*cos(x)^2 dx

becomes

∫ u^2 du

= (1/3)u^3

= (1/3)cos(x)^3

Putting everything together gives us

∫ [sin(x) - sin(x)*cos(x)^2] dx

**= -cos(x) + (1/3)cos(x)^3 + c**

## More Integral Calculus Help

- How to Integrate Sine Squared and Cosine Squared: sin(x)^2 and cos(x)^2
- How to Integrate sin(x)^4 and cos(x)^4, Fourth Power Trig Functions
- How to Integrate the Functions x*sin(x) and x*cos(x)
- How to Integrate Tan^2(x) and Tan^3(x), Tangent Squared and Cubed
- Two Ways to Integrate Sec(x), Secant of x

## Integrating cos(x)^3

In the same we we found the antiderivative of sin(x)^3, we can find the antiderivative of cos(x)^3. First we make the transformation

cos(x)^3 = cos(x) - cos(x)sin(x)^2

Now we make the u-substitution u = sin(x), du = cos(x). The integral

∫ [cos(x) - cos(x)*sin(x)^2] dx

becomes

∫ [1 - u^2] du

Evaluating this integral gives us

∫ [1 - u^2] du = u - (1/3)u^3

**= sin(x) - (1/3)sin(x)^3 + c**

## Graphs of Sin(x)^3 and Cos(x)^3 on the Same Plot

## Triple Angle Formula Alternatives

Another method of integrating sin(x)^3 and cos(x)^3 uses the triple angle identities

sin(x)^3 = 0.75*sin(x) - 0.25*sin(3x)

sin(3x) = 3*sin(x) - 4*sin(x)^3

cos(x)^3 = 0.75*cos(x) + 0.25*cos(3x)

cos(3x) = -3*cos(x) + 4*cos(x)^3

This produces the following integrals:

∫ sin(x)^3 dx

= ∫ [(3/4)sin(x) - (1/4)sin(3x)] dx**= -(3/4)cos(x) + (1/12)cos(3x) + c**

∫ cos(x)^3 dx

= ∫ [(3/4)cos(x) + (1/4)cos(3x)] dx**= (3/4)sin(x) + (1/12)sin(3x) + c**

As you might notice, these functional forms are different from the antiderivatives worked out in the previous two sections. However, they are completely equivalent. Just re-apply the triple angle substitution and you will recover the previous forms.

## Example Integration

A continuous probability distribution is given by the equation p(x) = K*cos(x)^3 over the interval [-π/2, π/2]. What should the value of K be so that the equation is a true probability distribution, i.e., the integral of p(x) over [-π/2, π/2] equals exactly 1?

To solve this problem, K must equal the reciprocal of ∫ cos(x)^3 dx evaluated from x = -π/2 to x = π/2. Working out the integral gives you

∫ cos(x)^3 dx

= ∫ [(3/4)cos(x) + (1/4)cos(3x)] dx

= (3/4)sin(x) + (1/12)sin(3x) + C

and

(3/4)sin(π/2) + (1/12)sin(3π/2) - (3/4)sin(-π/2) - (1/12)sin(-3π/2)

= 3/4 - 1/12 + 3/4 - 1/12

= 4/3

Therefore, K should be set equal to 3/4 and the probability distribution should be p(x) = (3/4)cos(x)^3.

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## Comments 1 comment

how is the triple angle identity derived? i cannot see it from the trig or geometry of a triangle with a trisected angle.