Integral of $$$\cos{\left(2 x \right)} \tan{\left(x \right)}$$$

Find $$$\int \cos{\left(2 x \right)} \tan{\left(x \right)}\, dx$$$.

Rewrite the integrand using the double angle formula for cosine $$$\cos{\left(2 x \right)}=2 \cos^{2}{\left(x \right)} - 1$$$:

$${\color{red}{\int{\cos{\left(2 x \right)} \tan{\left(x \right)} d x}}} = {\color{red}{\int{\left(2 \cos^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} d x}}}$$

Rewrite:

$${\color{red}{\int{\left(2 \cos^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} d x}}} = {\color{red}{\int{\left(2 \cos^{2}{\left(x \right)} \tan{\left(x \right)} - \tan{\left(x \right)}\right)d x}}}$$

Integrate term by term:

$${\color{red}{\int{\left(2 \cos^{2}{\left(x \right)} \tan{\left(x \right)} - \tan{\left(x \right)}\right)d x}}} = {\color{red}{\left(\int{2 \cos^{2}{\left(x \right)} \tan{\left(x \right)} d x} - \int{\tan{\left(x \right)} d x}\right)}}$$

Rewrite the tangent as $$$\tan\left(x\right)=\frac{\sin\left(x\right)}{\cos\left(x\right)}$$$:

$$\int{2 \cos^{2}{\left(x \right)} \tan{\left(x \right)} d x} - {\color{red}{\int{\tan{\left(x \right)} d x}}} = \int{2 \cos^{2}{\left(x \right)} \tan{\left(x \right)} d x} - {\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos{\left(x \right)}} d x}}}$$

Let $$$u=\cos{\left(x \right)}$$$.

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

The integral can be rewritten as

$$\int{2 \cos^{2}{\left(x \right)} \tan{\left(x \right)} d x} - {\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos{\left(x \right)}} d x}}} = \int{2 \cos^{2}{\left(x \right)} \tan{\left(x \right)} d x} - {\color{red}{\int{\left(- \frac{1}{u}\right)d u}}}$$

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

$$\int{2 \cos^{2}{\left(x \right)} \tan{\left(x \right)} d x} - {\color{red}{\int{\left(- \frac{1}{u}\right)d u}}} = \int{2 \cos^{2}{\left(x \right)} \tan{\left(x \right)} d x} - {\color{red}{\left(- \int{\frac{1}{u} d u}\right)}}$$

The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$\int{2 \cos^{2}{\left(x \right)} \tan{\left(x \right)} d x} + {\color{red}{\int{\frac{1}{u} d u}}} = \int{2 \cos^{2}{\left(x \right)} \tan{\left(x \right)} d x} + {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

Recall that $$$u=\cos{\left(x \right)}$$$:

$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} + \int{2 \cos^{2}{\left(x \right)} \tan{\left(x \right)} d x} = \ln{\left(\left|{{\color{red}{\cos{\left(x \right)}}}}\right| \right)} + \int{2 \cos^{2}{\left(x \right)} \tan{\left(x \right)} d x}$$

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

$$\ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\int{2 \cos^{2}{\left(x \right)} \tan{\left(x \right)} d x}}} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\left(2 \int{\cos^{2}{\left(x \right)} \tan{\left(x \right)} d x}\right)}}$$

Rewrite the integrand:

$$\ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + 2 {\color{red}{\int{\cos^{2}{\left(x \right)} \tan{\left(x \right)} d x}}} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + 2 {\color{red}{\int{\sin{\left(x \right)} \cos{\left(x \right)} d x}}}$$

Let $$$u=\sin{\left(x \right)}$$$.

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

The integral can be rewritten as

$$\ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + 2 {\color{red}{\int{\sin{\left(x \right)} \cos{\left(x \right)} d x}}} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + 2 {\color{red}{\int{u d u}}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:

$$\ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + 2 {\color{red}{\int{u d u}}}=\ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + 2 {\color{red}{\frac{u^{1 + 1}}{1 + 1}}}=\ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + 2 {\color{red}{\left(\frac{u^{2}}{2}\right)}}$$

Recall that $$$u=\sin{\left(x \right)}$$$:

$$\ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{u}}^{2} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\sin{\left(x \right)}}}^{2}$$

Therefore,

$$\int{\cos{\left(2 x \right)} \tan{\left(x \right)} d x} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + \sin^{2}{\left(x \right)}$$

Add the constant of integration:

$$\int{\cos{\left(2 x \right)} \tan{\left(x \right)} d x} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + \sin^{2}{\left(x \right)}+C$$

$$$\int \cos{\left(2 x \right)} \tan{\left(x \right)}\, dx = \left(\ln\left(\left|{\cos{\left(x \right)}}\right|\right) + \sin^{2}{\left(x \right)}\right) + C$$$A