Integrating Trigonometric Functions

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. 

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Integrating Trigonometric Functions Integrating Trigonometric Functions

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Contents
Table of contents
    • How do you integrate trigonometric functions?
      • Integral of sin(x)
      • Integral of cos(x)
      • Integral of tan(x)
    • 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 \(ln|\cos{x}|^{-1}\)

    \(|\cos{x}|^{-1} = \frac {1}{\cos{x}} = \sec {x}\) so \(\int{\tan{x} \space dx} = ln|\sec{x}| + c\)

    Find the integral of \(x \sin{2x}\)

    We will be using integration by parts, letting \(u = x\) since it will cancel out to \(\frac{du}{dx} = 1\).

    Therefore, \(dv = \sin {2x} \space dx\) and \(v = \frac {-\cos{2x}}{2}\), by the reverse chain rule.

    \(\begin{align} \int{x \sin {(2x)} \space dx} = \frac {-x}{2} \cos{(2x)} + \frac {1}{2} \int {\cos{(2x)} \space dx} \\ \frac {-x}{2} \cos{(2x)} + \frac {1}{4} \sin {(2x)} + c \end{align}\)

    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:

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

    Find \(\int{\cos^2{x} \space dx}\)

    We will use the identities \(\cos{2x} = \cos^2{x} - \sin^2{x}\) and \(\sin^2{x} = 1 - \cos^2{x}\)

    Rearranging these and combining them, we obtain \(\cos^2{x} = \frac{\cos{2x}}{2} + \frac {1}{2}\).

    We can then solve this integral.

    \(\begin{align} \int{\cos^2{x} \space dx} &= \frac {1}{2} \int {\cos{2x} + 1} \\ &= \frac {1}{2}(\frac{\sin{2x}}{2} + x) + c, \text {using the reverse Chain Rule for} \sin {2x} \\ &= \frac {\sin{2x}}{4} + \frac{x}{2} + c \end{align}\).

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

    Find \(\int{\arctan{2x} \space dx}\)

    We will need to use integration by substitution and by parts.

    Let a new variable t = 2x.

    Therefore, dt = 2 dx and \(\frac{dt}{2} = dx\).

    Substituting this into the integral, we get:

    \(\int{\arctan{t} \cdot \frac {dt}{2}} = \frac{1}{2} \int {\arctan{t} \cdot 1 \cdot dt}\)

    We will now use integration by parts, letting:

    \(u = \arctan{t}, \space du = \frac {1}{1 + t^2} dt, \space dv = 1dt, \space v = t\)

    Using the integration by parts formula, we get:

    \(\begin{align}\frac{1}{2} (t \cdot \arctan{t} - \int{\frac{t}{1 + t^2} dt} &= \frac{1}{2} t \cdot \arctan{t} - \frac{1}{2} \cdot \frac{1}{2} \int {\frac{2t}{1 + t^2} dt} \\ &= \frac{1}{2} t \cdot \arctan{t} - \frac{1}{4} ln|1 + t^2| \end{align}\).

    Since we let t = 2x, we now substitute x back in. Hence,

    Integrate\(\cos^3{x} \sin{x}\) with respect to x.

    We will use integration by substitution.

    \(\int{\cos^3{x} \sin{x} \space dx} = \int{(\cos{x})^3 \sin{x} \space dx}\).

    Letting \(u = \cos{x}, \space \frac{du}{dx} = -\sin{x}\) . Therefore, replacing u values by x values, we get \(\begin{align} \int{u^3(\frac{-du}{dx})dx} &= - \int{u^3du} \\ &= - \frac {u^4}{4} +c \end{align}\)

    We then replace the u values by the x values.

    Hence, \(\int{\cos^3{x} \sin{x} \space dx} = - \frac {\cos^4{x}}{4}+ c\)

    Integrating Trigonometric Functions Summary Table

    Trigonometric functionIntegral notationIntegral solution
    \(\sin{x}\)\(\int{\sin{x}}\space dx\)\(-\cos{x} + c\)
    \(\cos{x}\) \(\int{\cos{x} \space dx}\)\(\sin{x} + c\)
    \(\tan{x}\)\(\int{\tan{x} \space dx}\) \(ln|\cos{x}| + c\)
    \(\arcsin{x}\)

    \(\int {\arcsin{x} \space dx}\)

    \(x \cdot \arcsin{x} + \sqrt {1 - x^2} + c\).
    \(\arccos{x}\)\(\int{\arccos{x} \cdot dx}\)\(x \cdot \arccos{x} - \sqrt{1-x^2} + c\)
    \(\arctan{x}\)\(\int{\arctan{x} \space dx}\)\(x \space \arctan{x} - \frac{1}{2} ln|1 + x^2| + c\)

    Table 1. Integration of trigonometric functions.

    Integrating Trigonometric Functions - Key takeaways

    • \(\int{\sin{x} \space dx} = - \cos{x} + c\)
    • \(\int{\cos{x} \space dx} = \sin{x} + c\)
    • \(\int{\tan{x} \space dx} = ln|\sec{x}| + c\)
    • 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.
    Integrating Trigonometric Functions Integrating Trigonometric Functions
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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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