Integrating Trigonometric Functions

How do you integrate trigonometric functions?

Each trigonometric function has its defined integral:

Integral of sin(x)

The integral of \(\sin{x}\) is \(-\cos{x} + c\). Using integral notation, \(\int{\sin{x}}\space dx\).

Integral of cos(x)

The integral of \(\cos{x}\) is \(\sin{x} + c\) or \(\int{\cos{x}} dx = \sin {x} + c\).

Integral of tan(x)

The integral of tan(x) is \(ln|\cos{x}| + c\) or \(\int{\tan{x} dx} = ln|\cos{x}| + c\).

Let's look at the derivation of this.

We know that \(\tan{x} = \frac {\sin{x}}{\cos{x}}\) so we can substitute this into the integral \(\int{\tan{x} dx} = \int {\frac{\sin{x}}{\cos{x}}dx}\).

To solve this, we can use the substitution u = cos(x) so \(\frac{du}{dx} = -\sin{x}\) and \(dx = -\frac{1}{\sin{x}} du\).

Our integral will now look like this: \(\int{\frac{\sin{x}}{u}}{\frac{1}{-\sin{x}}} du\)

We can cancel out the \(\sin{x}\) and get \(\int{-\frac{1}{u} du}\).

We know that the integral of \( \frac{1}{x} = ln(x)\) , therefore \(\int{-\frac{1}{u} du} = -ln(u) + c\) .

If we substitute \(\cos{x}\) back in, we get \(\-ln \cdot \cos {x}\), which is equivalent to \(ln|\cos{x}|^{-1}\)

How do you integrate squared trigonometric functions?

To integrate squared trigonometric functions such as \(\sin^2{x}\), you can use the integrals for the trigonometric functions that you just determined, and double angle identities.

For example, to find \(\int{\sin^2{x} \space dx}\), you can use the identity \(\cos{2x} = 1 - 2\sin^2{x}\).

If we rearrange this expression to find \(\sin^2{x}\), you get \(\sin^2{x} = \frac{1}{2} - \frac {\cos{2x}}{2}\).

We can now substitute this into our integral:

\(\int{\sin^2{x} \space dx} = \int {\frac{1}{2} -\frac{\cos{2x}}{2} \space dx}\)

We know that the integral of \(\cos{x}\) is \(\sin{x}\) so the integral of \(\cos{2x}\) is \( \frac{1}{2} \sin {2x}\)

Taking the factor of \(\frac{1}{2}\) into account, we get:

\(\int{\sin^2{x} \space dx} = \frac {1}{2}x - \frac {1}{4} \sin {2x} + c\)

Integrating inverse trigonometric functions

Inverse trigonometric functions such as arcsin, arccos and arctan cannot be integrated directly. Therefore, we use Integration by Parts. We know that \(\int{u \space dv} = uv - \int {v \space du}\), and since we cannot integrate the inverse trigonometric function but we can derive it, we let u = inverse trigonometric function and v = 1. The integration by parts formula is then used to solve the integral.

Integral of arcsin(x)

The integral of \(\arcsin{x}\) can be written as \(\int{\arcsin{x} \cdot 1 \space dx}\).

Therefore, you let \(u = \arcsin {x}, du = \frac {1}{\sqrt{1-x^2}}, dv = 1, v =x\). .

We use the integration by parts formula and find the \(\int{\arcsin{x} \space dx} = x \cdot \arcsin {x} - \int {\frac {x}{\sqrt{1-x^2}} \space dx}\).

Let \(w = 1 - x^2\). Hence, \(dw = -2x \space dx\).

\(\int{\arcsin{x} \space dx} = x \cdot \arcsin {x} + \frac{1}{2} \int {-2x(1 - x^2)^{-\frac{1}{2}} \space dx}\).

Then, \(\int{\arcsin{x} \space dx} = x \cdot \arcsin{x} + \frac{1}{2} \cdot \frac {(1-x^2)^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} = x \cdot \arcsin{x} + (1 - x^2)^{\frac{1}{2}}\) .

Hence, \(\int {\arcsin{x} \space dx} = x \cdot \arcsin{x} + \sqrt {1 - x^2} + c\).

Integral of arccos(x)

The integral of \(\arccos{x}\) can be written as \(\int{\arccos{x} \cdot 1 \cdot dx}\). Using integration by parts, let \(u = \arccos{x}, du = \frac {-1}{\sqrt{1-x^2}}, dv = 1, v = x\) . Using the integration by parts formula, finding that \(\int{\arccos{x} \space dx} = x \cdot \arccos {x} - \int{\frac{-x}{\sqrt{1-x^2}} \space dx}\), or \(x \cdot \arccos{x} + \int{\frac{x}{\sqrt{1-x^2}} dx}\). We then use integration by substitution, letting \(w = 1 - x^2\).

Following the same method as for the integral of \(\arcsin{x}\), we find that \(\int{\arccos{x} \cdot dx} = x \cdot \arccos{x} - \sqrt{1-x^2} + c\).

Integral of arctan(x)

The integral of arctan(x) can be written as \(\int {\arctan{x} \cdot 1 \space dx}\). Using integration by parts, let \(u = \arctan{x}, \space du = \frac{1}{1 + x^2}, \space dv = 1, \space v = x\). Using the integration by parts formula, we find that \(\int\arctan{x} \space dx = x \cdot \arctan{x} - \int {\frac{x}{1 + x^2} dx}\). We recognize this integral as a natural logarithm of \((1 + x^2)\), since, letting \(w = 1 + x^2\), \(dw = 2x\). This means that the numerator \(x = \frac{1}{2} dw\).

We therefore find that \(\int{\arctan{x} \space dx} = x \space \arctan{x} - \frac{1}{2} ln|1 + x^2| + c\).

Integrating Trigonometric Functions Summary Table

Table 1. Integration of trigonometric functions.