Let's take a look at integrating trigonometric functions like sin, cos and tan as well as inverse trigonometric functions such as arcsin, arccos and arctan.
How do you integrate squared trigonometric functions?
Integrating inverse trigonometric functions
Integral of arcsin(x)
Integral of arccos(x)
Integral of arctan(x)
Integrating Trigonometric Functions Summary Table
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 \(\sin^2{x} = \frac{1}{2} - \frac {\cos{2x}}{2}\).
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 \(\int{\arcsin{x} \space dx} = x \cdot \arcsin {x} - \int {\frac {x}{\sqrt{1-x^2}} \space dx}\).
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 \(\int{\arctan{x} \space dx} = x \space \arctan{x} - \frac{1}{2} ln|1 + x^2| + c\).
We can use the chain rule when the variable in brackets is more complex than x, for example, \(\int{\sin{2x} \space dx = \frac {-1}{2} \cos{2x} + c\), as we have divided by the derivative of the brackets.
We can use and rearrange double angle identities, such as \(\cos{2x} = 2 \cos^2{x} - 1\) when given a squared trigonometric function.
When calculating integrals of inverse trigonometric functions, we use integration by parts, using the formula \(int{u \space dv} = uv - \int{v \space du}\), and letting u = inverse trigonometric function, and dv = 1.
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Frequently Asked Questions about Integrating Trigonometric Functions
How do you integrate trigonometric functions?
The integral of sinx is -cosx + c. The integral of cosx is sinx + c. The integral of tanx is ln(secx) + c. Use the reverse chain rule if the variable is more complex than x. Integration by parts may also be required, for example to find the integral of xsinx with respect to x.
How do you integrate inverse trigonometric functions?
Let the inverse trigonometric function = u and let dv = 1. Hence, find du and v, and use the integration by parts formula to solve. The reverse chain rule may be needed if the variable is more complex than x.
How do you integrate squared trigonometric functions?
By using double angle identities, such as cos2x=1-2(sin^2(x)). By rearranging the expression to find sin^2(x), for example, its equivalent can be subbed in to solve the integral. In these cases, the reverse chain rule is often required.
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