# How to Integrate sin^5(x) and cos^5(x) | sin(x)^5 and cos(x)^5

TR Smith is a product designer and former teacher who uses math in her work every day.

Graph of y = sin(x)^5 in red, graph of y = 2 + cos(x)^5 in blue

The graphs above show the functions sin(x) raised to the 5th power and cos(x) raised to the 5th power. These functions can be written either as f(x) = sin(x)^5 and g(x) = cos(x)^5, or as f(x) = sin^5(x) and g(x) = cos^5(x). To integrate them, you only need to use the identity sin(x)^2 + cos(x)^2 = 1 and the substitution rule of integral calculus.

The antiderivates of any odd powers of sine and cosine can be determined using these transformations. (See for example the antiderivative of sin(x)^3.) However, the even powers of sine and cosine are more difficult to work out and require more advanced techniques with trigonometric identities. (See for example the antiderivative of sin(x)^4.)

This tutorial covers the integration of sin(x)^5 and cos(x)^5 with some example problems.

## Integral of sin(x)^5 or sin^5(x)

To work out the integral of sin(x)^5, we first write it as sin(x)*sin(x)^4 = sin(x)*[sin(x)^2]^2. Why use this equivalent form? Because sin(x)^2 can be replaced by 1 - cos(x)^2. This gives us

∫ [1 - cos(x)^2]^2 * sin(x) dx

Now we can use the substitution u = cos(x) and du = -sin(x) dx. This transforms our integral into

∫ -[1 - u^2]^2 du
= ∫ [-1 + 2u^2 - u^4] du
= -u + (2/3)u^3 - (1/5)u^5 + C

Making the reverse substitution u = cos(x) gives us

∫ sin(x)^5 dx = -cos(x) + (2/3)cos(x)^3 - (1/5)cos(x)^5 + C

## Integral of cos^5(x) or cos(x)^5

This integral is worked out almost exactly the same way as the previous antiderivative. First we rewrite cos(x)^5 in the equivalent form cos(x)*[cos(x)^2]^2 = cos(x)[1 - sin(x)^2]^2. This gives us

∫ cos(x)^5 dx = ∫ [1 - sin(x)^2]^2 * cos(x) dx

The substitution we make now is v = sin(x) and dv = cos(x) dx. This gives us the polynomial integral

∫ [1 - v^2]^2 dv
= ∫ [1 - 2v^2 + v^4] dv
= v - (2/3)v^3 + (1/5)v^5 + C

And now with the reverse substitution v = sin(x) we finally obtain

∫ cos(x)^5 dx = sin(x) - (2/3)sin(x)^3 + (1/5)sin(x)^5 + C

This is almost the same expression as the previous integral, except multiplied by -1 and with sines instead of cosines.

## How to Use the Antiderivative of sin(x)^5 : Example

The graph above shows y = sin(x)^5 from x = 0 to x = π, with the region above the x-axis shaded. What is the area under the curve?

To find the area under the curve and above the axis line for y = sin(x)^5, you simply integrate the function from x = 0 to x = π.

∫ sin(x)^5 dx, {0 ≤ x ≤ π}
= -cos(x) + (2/3)cos(x)^3 - (1/5)cos(x)^5, {0 ≤ x ≤ π}

To evaluate the integral over the given interval, we plug in the endpoints 0 and π and subtract. This gives us

[-cos(π) + (2/3)cos(π)^3 - (1/5)cos(π)^5] -
[-cos(0) + (2/3)cos(0)^3 - (1/5)cos(0)^5]

= [-(-1) + (2/3)(-1) - (1/5)(-1)] - [-1 + (2/3)(1) - (1/5)(1)]

= 16/15

## How to Use the Antiderivative of cos(x)^5 : Example

Let J(t) = (15/16)cos(t)^5 be a probability density function for -π/2 ≤ t ≤ π/2. What is the value of 'a' such that P(-a ≤ t ≤ a) = 0.5? I.e., what is the value of 'a' such that the probability that the random variable t is between -a and a equals exactly 50%?

To solve this problem, we need to solve the integral equation

0.5 = ∫ (15/16)cos(t)^5 dt, {-a ≤ t ≤ a}

8/15 = ∫ cos(t)^5 dt, {-a ≤ t ≤ a}

Taking the antiderivative and evaluating at the endpoints t = a and t = -a we get

8/15 = 2sin(a) - (4/3)sin(a)^3 + (2/5)sin(a)^5

This is obtained using the fact that sin(a) = -sin(-a). If we replace 'sin(a)' with 'z' and clear the denominators, we get the quintic polynomial equation

2z^5 - 20z^3 + 30z - 8 = 0

z^5 - 10z^3 + 15z - 4 = 0

Using a solver tool such as the TI-84 or a quintic equation solver, we get the following solutions for z:

z ≈ 2.89375
z ≈ 1.15329
z ≈ 0.281405
z ≈ -1.51258
z ≈ -2.81586

Since z = sin(a) and the sine of a number must be between -1 and 1, the only value of z that works is z ≈ 0.281405. This means that a ≈ arcsin(0.281405) ≈ 0.285. So the probability that a random variable t (with a distribution of J(t)) lies between -0.285 and 0.285 is 50%.

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• km 2 years ago

What is the antiderivative of (sin(5x)^5)cos(5x)?

• Author

TR Smith 2 years ago from Eastern Europe

Use the substitution u = sin(5x) and du = 5cos(5x) dx. Then you'll end up with the simple polynomial integral (1/5)u^5, whose antiderivative is (1/30)u^6 + c. Therefore the antiderivative of (sin(5x)^5)cos(5x) is (1/30)sin(5x)^6 + c.