Example 5. Find $\int \frac{d x}{1+\sin ^{2} x}$.
Solution. There is no factor that would give the differential of some function (new variable). This means we need to obtain it.

Method 1. The presence of $\sin ^{2} x$ in the denominator intuitively leads to the desire to use the relation $\frac{d x}{\sin ^{2} x}=-d(\operatorname{ctg} x)$. How can we get this fraction? Apparently, like this: $\frac{1}{1+\sin ^{2} x}=\frac{1}{\sin ^{2} x \cdot\left(\frac{1}{\sin ^{2} x}+1\right)}$ (we factored out $\sin ^{2} x$). It remains to express $\frac{1}{\sin ^{2} x}+1$ in terms of $\operatorname{ctg} x$. This is possible because $1+\operatorname{ctg}^{2} x=\frac{1}{\sin ^{2} x}$. We implement the desired as follows:

$$
\begin{aligned}
& \int \frac{d x}{1+\sin ^{2} x}=\int \frac{1}{\frac{1}{\sin ^{2} x}+1} \cdot \frac{d x}{\sin ^{2} x}= \\
&-\int \frac{1}{2+\operatorname{ctg}^{2} x} d(\operatorname{ctg} x)=\{u=\operatorname{ctg} x\}=-\int \frac{d u}{2+u^{2}}= \\
&=\left\{\begin{array}{l}
\text { we have arrived at a tabular formula 12, } \\
\text { which can be expressed using arccotangent }
\end{array}\right\}= \\
& \quad=\frac{1}{\sqrt{2}} \operatorname{arcctg} \frac{u}{\sqrt{2}}+C=\frac{1}{\sqrt{2}} \operatorname{arctg} \frac{\operatorname{ctg} x}{\sqrt{2}}+C
\end{aligned}
$$

Method 2 is based on a similar idea, implemented using the tangent (first, we replace 1 with $\sin ^{2} x+\cos ^{2} x$):

$$
\begin{aligned}
\int \frac{d x}{1+\sin ^{2} x}=\int \frac{d x}{2 \sin ^{2} x+\cos ^{2} x} & =\int \frac{1}{2 \operatorname{tg}^{2} x+1} \cdot \frac{d x}{\cos ^{2} x}= \\
& =\int \frac{d(\operatorname{tg} x)}{2 \operatorname{tg}^{2} x+1}=\frac{1}{\sqrt{2}} \operatorname{arctg}(\sqrt{2} \operatorname{tg} x)+C .
\end{aligned}
$$

Here, the intermediate integral $\int \frac{d u}{(\sqrt{2} u)^{2}+1}=$ $=\frac{1}{\sqrt{2}} \operatorname{arctg} \sqrt{2} u+C$ (according to formula (*)) is omitted. Note that methods 1 and 2 are equivalent.

$$
\text { Example 6. Find } \int \frac{d x}{\sin x} \text {. }
$$

Solution. This integral is included in the table only for its completeness, although, unlike other tabular integrals, it is not trivial. The method of obtaining this result is instructive because it is associated with obtaining a factor for inclusion under the differential sign.

Method 1 is based on the expansion of the sine of a double argument followed by transformation to obtain the required factor. We have:

$$
\int \frac{d x}{\sin x}=\int \frac{d x}{2 \sin \frac{x}{2} \cos \frac{x}{2}}=\int \frac{1}{\operatorname{tg} \frac{x}{2}} \cdot \frac{d x}{2 \cos ^{2} \frac{x}{2}}=\int \frac{d\left(\operatorname{tg} \frac{x}{2}\right)}{\operatorname{tg} \frac{x}{2}}=\ln \left|\operatorname{tg} \frac{x}{2}\right|+C
$$

Here, we have two levels of complexity of the function to be included under the differential sign: the tangent and the half-argument.

Method 2 is based on the use of the trigonometric identity:

$$
\begin{aligned}
\int \frac{d x}{\sin x}= & \int \frac{\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}} d x=\{\text { decompose into two fractions }\}= \\
= & \int \frac{\sin \frac{x}{2}}{2 \cos \frac{x}{2}} d x+\int \frac{\cos \frac{x}{2}}{2 \sin \frac{x}{2}} d x=-\int \frac{d\left(\cos \frac{x}{2}\right)}{\cos \frac{x}{2}}+\int \frac{d\left(\sin \frac{x}{2}\right)}{\sin \frac{x}{2}}= \\
& -\ln \left|\cos \frac{x}{2}\right|+\ln \left|\sin \frac{x}{2}\right|+C=\ln \left|\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right|+C=\ln \left|\operatorname{tg} \frac{x}{2}\right|+C .
\end{aligned}
$$

This method may be more elegant, although longer.