We use trigonometric functions quite often in integration, even when there are no trig. functions explicitly in the integral. In *this* section we'll look at integrals that contain only trigonometric functions, such as

$$ \begin{align} \int sin^3(&x) \, dx \\[5pt] \int sin^2(&x) tan(x) \, dx \\[5pt] \int cos^4(&x) sin^2(x) \, dx \end{align}$$

Trigonometric integrals like these are very important throughout science and engineering. The field of **Fourier analysis**, which includes **fourier series** and the "**fourier transform**", is based upon series of terms that include trig. functions,

and we often have to integrate complicated trigonometric expressions. You couldn't get an MRI image of your aching knee, for example, if it weren't for the Fourier transform and trigonometric integrals.

We'll develop some tricks for making these complicated integrals possible. This section will build heavily on your knowledge of analytical trigonometry (trig. identities), integration by u-substitution and integration by parts.

In another section, we'll cover how it's possible to integrate some pretty tricky functions by substituting trig. functions for variables, taking advantage of some of the special algebra of trig functions, such as the Pythagorean identity,

$$\bf sin^2(x) + cos^2(x) = 1$$

First we recall the **Pythagorean identity**:

$$sin^2(x) + cos^2(x) = 1$$

If we begin with the cosine double angle formula (derived here),

$$cos(2\theta) = cos^2(\theta) - sin^2(\theta)$$

we can use the Pythagorean identity to substitute 1 - cos^{2}θ for sin^{2}θ to obtain one of the **power-reduction identities**:

Notice that this identity allows us down-convert the power of the cosine function from 2 to 1. And it's easy to integrate a function like cos(2θ) or sin(2θ) by simple u-substitution.

This will be very valuable in doing trig. integrals. Likewise, we can develop a **sine-power conversion formula**:

In a similar way, we can take the Pythagorean identity for secant and tangent,

$$sec^2(\theta) = 1 + tan^2(\theta)$$

(obtained by dividing the Pythagorean identity through by cos^{2}θ), to obtain the tangent power-reduction formula:

$$tan^2(\theta) = 1 - \frac{2 tan(\theta)}{tan(2 \theta)}$$

Here is a recap of the most important trig. identities we'll use to do trigonometric integrals:

In the examples that follow, we'll first do integrals in which at least one of the exponents of a trig. function is *odd*. These will turn out only to need the Pythagorean identities. When all exponents of trig. functions are *even*, we'll need to use the power-reduction formulae to *create* an odd exponent, then proceed as before.

**Solution****sin ^{3}x = sin(x)·sin^{2}(x)**, and write an equivalent integral:

$$\int sin(x) sin^2(x) \, dx$$

Now use the Pythagorean identity

$$sin^2(x) = 1 - cos^2(x)$$

to replace **sin ^{2}x** and rewrite the integral:

$$\int sin(x)(1 - cos^2(x)) \, dx$$

Now if we distribute the **sin(x)** into the parenthesis, this new integral is a sum of two integrals, the last of which can be evaluated easily using the substitution **u = cos(x)**, like this:

The first integral is easy, it's just **-cos(x)**. The second is relatively easy because of the substitution.

Now we just back substitute **cos(x)** for **u** to get the solution (don't forget the constant).

$$\int sin^3(x) \, dx = \frac{1}{3} cos^3 (x) - cos (x) + C$$

*because* the coefficient of the sine function was **odd**. In order for a substitution like this to work in a trigonometric integral, at least one of the exponents of a trig. function must be **odd**.

**Solution**

From this, multiplying through by tan(x), we get two integrals (We'll call them **1** and **2**), each of which can be solved by simple substitution (u-substitution). Here's what it looks like:

**Integral 1**: The first integral suggests a substitution because tan(x) and sec(x) are related by a derivative.

$$\int tan(x) sec^2(x) \, dx$$

One substitution is

let u = sec(x),

then du = sec(x)tan(x) dx

Then our new integral can be done very simply:

$$\int u\, du \ \frac{u^2}{2} + C$$

Re-substitution of sec(x) for u gives:

$$\int tan(x) sec^2(x) \, dx = \frac{1}{2} sec^2(x) + C$$

**Integral 2**: The second integral is one you might just remember, but in case you've forgotten, expand tangent into sine/cosine:

$$\int tan(x) \, dx = \int \frac{sin(x)}{cos(x)} \, dx$$

Then make the substitution

let u = cos(x),

then du = -sin(x) dx

That gives us a new integral that's easy to solve:

$$-\int \frac{du}{u} = -|ln|u| + C$$

Back-substituting for u gives

$$\int tan(x) \, dx = -ln|cos(x)| + C$$

Putting both parts together gives us the solution:

$$\int tan^3(x) \, dx = \frac{1}{2} sec^2(x) + ln|cos(x)| + C$$

**Solution****sin(x)**, which will come in handy later.

$$\int sin(x) (1 - cos^2(x)) cos^2(x) \, dx$$

Expand the integral to get the integral of a sum.

$$\int [sin(x) cos^2(x) - sin(x) cos^4(x)] \, dx$$

Each of these is now easy to do with the *same* simple substitution:

let u = cos(x), then du = -sin(x) dx

to get these integrals:

$$-\int u^2 \, du + \int u^4 \, du$$

Evaluation of these and re-substitution of **cos(x)** for **u** gives the final integral:

$$-\frac{1}{3} cos^3(x) + \frac{1}{5} cos^5(x) + C$$

The only thing that remains is to do an example in which the powers of the trig. functions are higher.

**Solution**

$$\int sin^4(x) cos^5(x) \, dx =$$

We use the Pythagorean identity for sine and cosine to re-express cos^{4}(x) in terms of sines:

$$= \int sin^4(x) (1 - sin^2(x))^2 cos(x) \, dx$$

Expanding the binomial (1 - sin^{2}(x))^{2} gives:

$$= \int sin^4(x) [1 - 2 sin^2(x) + sin^4(x)] cos(x) dx$$

and multiplying through by sin4(x) gives us three separate terms which can be integrated separately.

$$ \begin{align} = \int sin^4(&x) cos(x) \, dx \\ &- 2 \int sin^6(x) cos(x) \, dx \\ &+ \int sin^8(x) cos(x) \, dx \end{align}$$

But now each of our terms contains cos(x), whcih we can make use of in a u-substitution:

let u = sin(x), then du = cos(x) dx

Notice that cos(x) dx is in each of our integrals, so the substitution gives integrals that are easy to evaluate:

$$ \begin{align} &\int u^4 \, du - 2 \int u^6 \, du + \int u^8 \, du \\ \\ &= \frac{1}{5} u^5 = \frac{2}{7} u^7 + \frac{1}{9} u^9 + C \end{align}$$

Finally, we can substitute sin(x) back in for u to get the result:

$$ \begin{align} \int &sin^4(x) cos^5(x) \, dx \\ &= \frac{1}{5} sin^5(x) - \frac{2}{7} sin^7(x) + \frac{1}{9} sin^9(x) + C \end{align}$$

Just so we don't leave the tangent function out, we'll do another example below with a mixed tangent-secant integrand.

**Solution****tan ^{2}(x)** with

$$\int tan(x) (sec^2(x) - 1) sec^2(x) \, dx$$

Now expand to get the integral of a sum (which is the sum of integrals):

$$\int [tan(x) sec^4(x) - tan(x) sec^2(x) \, dx$$

Then make the same substitution in both,

let u = sec(x),

then du = sec(x)tan(x) dx

and solve the easy power integrals:

$$\int u^3 \, du - \int u \, du = \frac{1}{4} u^4 - \frac{1}{2} u^2 + C$$

Finally, back substitute to get the solution.

$$ \begin{align} \int tan^3(x) &sec^2(x) \, dx \\ &= \frac{1}{4} sec^4(x) - \frac{1}{2} sec^2(x) + C \end{align}$$

**Solution**

$$sin^2(x) = \frac{1 - cos(2x)}{2}$$

to make the simple substitution

$$int sin^2(x) \, dx = \frac{1}{2} \int [1 - cos(2x)] \, dx$$

Then we can separate this integral of a sum into the sum of integrals. The first is trivial, and the second can be don by u-substitution.

$$= \frac{1}{2} \left[ \int dx - \int cos(2x) \, dx \right]$$

Both integrals are easy now (the first is already done below). The second can be done with the substitution

giving the integral

$$\frac{x}{2} - \frac{1}{4} \int cos(u) \, du$$

We then take the result,

$$= \frac{x}{2} - \frac{1}{4} sin(u) + C$$

and back substitute **u = 2x** to arrive at the solution to the integral:

$$\int sin^2(x) \, dx = \frac{x}{2} - \frac{1}{4} sin(2x) + C$$

**Solution**

to get a new integral:

$$ \begin{align} \frac{1}{4} \int (1 - &cos(2x))(1 + cos(2x) \, dx) \\ &= \frac{1}{4} \int (1 - cos^2(2x)) \, dx \end{align}$$

Now this integral contains a **cos ^{2}** term. We can use the power reduction formula again here, this time with

$$cos^2(2x) = \frac{1 + cos(4x)}{2}$$

Make the substitution and carry out the integration. To integrate the last term, we use the u-substitution **u = 4x**, then **du = 4dx**. Be careful with all of the various fractional factors that result.

Finally, the integral is

$$\int sin^2(x) cos^2(x) dx = \frac{x}{8} + \frac{1}{32} sin(4x) + C$$

That should be enough examples for you to try some on your own.

Find the following indefinite integrals:

1. | $$\int sin^4(x) \, dx$$ | |

2. | $$\int cos^2(x) tan^3(x) \, dx$$ | |

3. | $$\int sin^2(x) cos^4(x) \, dx$$ |

4. | $$\int tan^3(x) \, dx$$ | |

5. | $$\int cos^2(x) tan^3(x) sec^2(x) \, dx$$ | |

6. | $$\int sin^5(x) cos^2(x) \, dx$$ |

Itegration of **tan(x)·cos ^{2}(x) dx**. Requires integration by parts.

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