Integral of $$$\frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}}$$$

The calculator will find the integral/antiderivative of $$$\frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}}$$$, 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 contact us.

Your Input

Find $$$\int \frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}}\, dx$$$.

Solution

Let $$$u=8 x$$$.

Then $$$du=\left(8 x\right)^{\prime }dx = 8 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{8}$$$.

Thus,

$${\color{red}{\int{\frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}} d x}}} = {\color{red}{\int{\frac{\sin{\left(2 u \right)}}{8 \cos{\left(u \right)}} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{8}$$$ and $$$f{\left(u \right)} = \frac{\sin{\left(2 u \right)}}{\cos{\left(u \right)}}$$$:

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

Rewrite the integrand:

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

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

$$\frac{{\color{red}{\int{2 \sin{\left(u \right)} d u}}}}{8} = \frac{{\color{red}{\left(2 \int{\sin{\left(u \right)} d u}\right)}}}{8}$$

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

$$\frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{4} = \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{4}$$

Recall that $$$u=8 x$$$:

$$- \frac{\cos{\left({\color{red}{u}} \right)}}{4} = - \frac{\cos{\left({\color{red}{\left(8 x\right)}} \right)}}{4}$$

Therefore,

$$\int{\frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}} d x} = - \frac{\cos{\left(8 x \right)}}{4}$$

Add the constant of integration:

$$\int{\frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}} d x} = - \frac{\cos{\left(8 x \right)}}{4}+C$$

Answer

$$$\int \frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}}\, dx = - \frac{\cos{\left(8 x \right)}}{4} + C$$$A


Please try a new game Rotatly