Integral of $$$\frac{1 - \sin^{2}{\left(x \right)}}{\cos^{2}{\left(x \right)}}$$$

Find $$$\int \frac{1 - \sin^{2}{\left(x \right)}}{\cos^{2}{\left(x \right)}}\, dx$$$.

Expand the expression:

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

Integrate term by term:

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

Rewrite the integrand in terms of the secant:

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

The integral of $$$\sec^{2}{\left(x \right)}$$$ is $$$\int{\sec^{2}{\left(x \right)} d x} = \tan{\left(x \right)}$$$:

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

Rewrite in terms of the tangent:

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

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

Then $$$x=\operatorname{atan}{\left(u \right)}$$$ and $$$dx=\left(\operatorname{atan}{\left(u \right)}\right)^{\prime }du = \frac{du}{u^{2} + 1}$$$ (steps can be seen »).

Thus,

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

Rewrite and split the fraction:

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

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

The integral of $$$\frac{1}{u^{2} + 1}$$$ is $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:

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

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

$$\tan{\left(x \right)} + \operatorname{atan}{\left({\color{red}{u}} \right)} - {\color{red}{u}} = \tan{\left(x \right)} + \operatorname{atan}{\left({\color{red}{\tan{\left(x \right)}}} \right)} - {\color{red}{\tan{\left(x \right)}}}$$

Therefore,

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

Simplify:

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

Add the constant of integration:

$$\int{\frac{1 - \sin^{2}{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x} = x+C$$

$$$\int \frac{1 - \sin^{2}{\left(x \right)}}{\cos^{2}{\left(x \right)}}\, dx = x + C$$$A