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