Integral of $$$\cos^{4}{\left(x \right)}$$$

The calculator will find the integral/antiderivative of $$$\cos^{4}{\left(x \right)}$$$, with steps shown.

Related calculator: Definite and Improper Integral Calculator

Please write without any differentials such as $$$dx$$$, $$$dy$$$ etc.
Leave empty for autodetection.

If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.

Your Input

Find $$$\int \cos^{4}{\left(x \right)}\, dx$$$.

Solution

Rewrite the cosine using the power reducing formula $$$\cos^{4}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{\cos{\left(4 \alpha \right)}}{8} + \frac{3}{8}$$$ with $$$\alpha=x$$$:

$${\color{red}{\int{\cos^{4}{\left(x \right)} d x}}} = {\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{\cos{\left(4 x \right)}}{8} + \frac{3}{8}\right)d x}}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{8}$$$ and $$$f{\left(x \right)} = 4 \cos{\left(2 x \right)} + \cos{\left(4 x \right)} + 3$$$:

$${\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{\cos{\left(4 x \right)}}{8} + \frac{3}{8}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(4 \cos{\left(2 x \right)} + \cos{\left(4 x \right)} + 3\right)d x}}{8}\right)}}$$

Integrate term by term:

$$\frac{{\color{red}{\int{\left(4 \cos{\left(2 x \right)} + \cos{\left(4 x \right)} + 3\right)d x}}}}{8} = \frac{{\color{red}{\left(\int{3 d x} + \int{4 \cos{\left(2 x \right)} d x} + \int{\cos{\left(4 x \right)} d x}\right)}}}{8}$$

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=3$$$:

$$\frac{\int{4 \cos{\left(2 x \right)} d x}}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{{\color{red}{\int{3 d x}}}}{8} = \frac{\int{4 \cos{\left(2 x \right)} d x}}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{{\color{red}{\left(3 x\right)}}}{8}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=4$$$ and $$$f{\left(x \right)} = \cos{\left(2 x \right)}$$$:

$$\frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{{\color{red}{\int{4 \cos{\left(2 x \right)} d x}}}}{8} = \frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{{\color{red}{\left(4 \int{\cos{\left(2 x \right)} d x}\right)}}}{8}$$

Let $$$u=2 x$$$.

Then $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (steps can be seen here), and we have that $$$dx = \frac{du}{2}$$$.

Therefore,

$$\frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{{\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{2} = \frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{2}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$\frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{2} = \frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{2}$$

The integral of the cosine is $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

$$\frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{4} = \frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{{\color{red}{\sin{\left(u \right)}}}}{4}$$

Recall that $$$u=2 x$$$:

$$\frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{\sin{\left({\color{red}{u}} \right)}}{4} = \frac{3 x}{8} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{\sin{\left({\color{red}{\left(2 x\right)}} \right)}}{4}$$

Let $$$u=4 x$$$.

Then $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (steps can be seen here), and we have that $$$dx = \frac{du}{4}$$$.

Thus,

$$\frac{3 x}{8} + \frac{\sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\cos{\left(4 x \right)} d x}}}}{8} = \frac{3 x}{8} + \frac{\sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{8}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$\frac{3 x}{8} + \frac{\sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{8} = \frac{3 x}{8} + \frac{\sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}}{8}$$

The integral of the cosine is $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

$$\frac{3 x}{8} + \frac{\sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{32} = \frac{3 x}{8} + \frac{\sin{\left(2 x \right)}}{4} + \frac{{\color{red}{\sin{\left(u \right)}}}}{32}$$

Recall that $$$u=4 x$$$:

$$\frac{3 x}{8} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left({\color{red}{u}} \right)}}{32} = \frac{3 x}{8} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left({\color{red}{\left(4 x\right)}} \right)}}{32}$$

Therefore,

$$\int{\cos^{4}{\left(x \right)} d x} = \frac{3 x}{8} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(4 x \right)}}{32}$$

Simplify:

$$\int{\cos^{4}{\left(x \right)} d x} = \frac{12 x + 8 \sin{\left(2 x \right)} + \sin{\left(4 x \right)}}{32}$$

Add the constant of integration:

$$\int{\cos^{4}{\left(x \right)} d x} = \frac{12 x + 8 \sin{\left(2 x \right)} + \sin{\left(4 x \right)}}{32}+C$$

Answer: $$$\int{\cos^{4}{\left(x \right)} d x}=\frac{12 x + 8 \sin{\left(2 x \right)} + \sin{\left(4 x \right)}}{32}+C$$$