Integral of $$$\tan{\left(x \right)} - \sec{\left(x \right)}$$$

Find $$$\int \left(\tan{\left(x \right)} - \sec{\left(x \right)}\right)\, dx$$$.

Integrate term by term:

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

Rewrite the secant as $$$\sec\left(x\right)=\frac{1}{\cos\left(x\right)}$$$:

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

Rewrite the cosine in terms of the sine using the formula $$$\cos\left(x\right)=\sin\left(x + \frac{\pi}{2}\right)$$$ and then rewrite the sine using the double angle formula $$$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$$:

$$\int{\tan{\left(x \right)} d x} - {\color{red}{\int{\frac{1}{\cos{\left(x \right)}} d x}}} = \int{\tan{\left(x \right)} d x} - {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}}$$

Multiply the numerator and denominator by $$$\sec^2\left(\frac{x}{2} + \frac{\pi}{4} \right)$$$:

$$\int{\tan{\left(x \right)} d x} - {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}} = \int{\tan{\left(x \right)} d x} - {\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}}$$

Let $$$u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}$$$.

Then $$$du=\left(\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}\right)^{\prime }dx = \frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2} dx$$$ (steps can be seen »), and we have that $$$\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)} dx = 2 du$$$.

Therefore,

$$\int{\tan{\left(x \right)} d x} - {\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}} = \int{\tan{\left(x \right)} d x} - {\color{red}{\int{\frac{1}{u} d u}}}$$

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

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

Recall that $$$u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}$$$:

$$- \ln{\left(\left|{{\color{red}{u}}}\right| \right)} + \int{\tan{\left(x \right)} d x} = - \ln{\left(\left|{{\color{red}{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}}}\right| \right)} + \int{\tan{\left(x \right)} d x}$$

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

$$- \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)} + {\color{red}{\int{\tan{\left(x \right)} d x}}} = - \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)} + {\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos{\left(x \right)}} d x}}}$$

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 $$$\sin{\left(x \right)} dx = - du$$$.

Thus,

$$- \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)} + {\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos{\left(x \right)}} d x}}} = - \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)} + {\color{red}{\int{\left(- \frac{1}{u}\right)d u}}}$$

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

$$- \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)} + {\color{red}{\int{\left(- \frac{1}{u}\right)d u}}} = - \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)} + {\color{red}{\left(- \int{\frac{1}{u} d u}\right)}}$$

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

$$- \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)} - {\color{red}{\int{\frac{1}{u} d u}}} = - \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)} - {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

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

$$- \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)} - \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = - \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)} - \ln{\left(\left|{{\color{red}{\cos{\left(x \right)}}}}\right| \right)}$$

Therefore,

$$\int{\left(\tan{\left(x \right)} - \sec{\left(x \right)}\right)d x} = - \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} - \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)}$$

Add the constant of integration:

$$\int{\left(\tan{\left(x \right)} - \sec{\left(x \right)}\right)d x} = - \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} - \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)}+C$$

$$$\int \left(\tan{\left(x \right)} - \sec{\left(x \right)}\right)\, dx = \left(- \ln\left(\left|{\cos{\left(x \right)}}\right|\right) - \ln\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right|\right)\right) + C$$$A