1037. Find the integrals:
1) $\int \frac{d x}{2 x^{2}+9}$
2) $\int \frac{d x}{\sqrt{5-4 x^{2}}}$
3) $\int \frac{d x}{\sqrt{2 x^{2}-3}}$
4) $\int \frac{d x}{4 x^{2}-3}$
5) $\int \cos ^{2} \frac{x}{2} d x$
6) $\int \operatorname{tg}^{2} x d x$
Solution. 1) We will transform the given integral to a tabular form as follows:

$$
I=\int \frac{d x}{2 x^{2}+9}=\int \frac{d x}{2\left(x^{2}+\frac{9}{2}\right)}=\frac{1}{2} \int \frac{d x}{x^{2}+\frac{9}{2}}
$$

here $a^{2}=\frac{9}{2}$ (see formula 12), therefore,

$$
I=\frac{1}{2} \cdot \frac{\frac{1}{3}}{\sqrt{\overline{2}}} \operatorname{arctg} \frac{x}{\frac{3}{\sqrt{2}}}+C=-\frac{\sqrt{2}}{6} \operatorname{arctg} \frac{x \sqrt{2}}{3}+C .
$$

We will perform similar transformations for integrals 2, 3, and 4.

$$
\begin{aligned}
& \text { 2) } \int \frac{d x}{\sqrt{5-4 x^{2}}}=\int \frac{d x}{\sqrt{4\left(\frac{5}{4}-x^{2}\right)}}=\frac{1}{2} \int \frac{d x}{\sqrt{\frac{5}{4}-x^{2}}}= \\
& =\frac{1}{2} \arcsin \frac{2 x}{\sqrt{5}}+C \quad[(11)]
\end{aligned}
$$

The answer can also be written in another form. Transforming,

$\frac{1}{\sqrt{2}} \ln \left|x+\sqrt{x^{2}-\frac{3}{2}}\right|+C=\frac{1}{\sqrt{2}} \ln \left|\frac{x \sqrt{2}+\sqrt{2 x^{2}-3}}{\sqrt{2}}\right|+C=$ $=\frac{1}{\sqrt{2}} \ln \left|x \sqrt{2}+\sqrt{2 x^{2}-3}\right|+C_{1}$.

Thus,

$$
\begin{aligned}
& \int \frac{d x}{\sqrt{2 x^{2}-3}}=\frac{1}{\sqrt{2}} \ln \left|x+\sqrt{x^{2}-\frac{3}{2}}\right|+C= \\
& \quad=\frac{1}{\sqrt{2}} \ln \left|x \sqrt{2}+\sqrt{2 x^{2}-3}\right|+C_{1}
\end{aligned}
$$

Both answers are equivalent, which can be verified by differentiation.

$$
\begin{aligned}
& \text { 4) } \int \frac{d x}{4 x^{2}-3}=\frac{1}{4} \int \frac{d x}{x^{2}-\frac{3}{4}}=\frac{1}{4} \cdot \frac{1}{2 \sqrt{\frac{3}{4}}} \ln \left|\frac{x-\sqrt{\frac{3}{4}}}{x+\sqrt{\frac{3}{4}}}\right|+C= \\
= & \frac{1}{4 \sqrt{3}} \ln \left|\frac{2 x-\sqrt{3}}{2 x+\sqrt{3}}\right|+C \quad[(13)] .
\end{aligned}
$$

5) The given integral is not a tabular one. We will represent the integrand as an algebraic sum and then apply the formulas and rules of integration:

$$
\begin{aligned}
\int \cos ^{2} \frac{x}{2} d x & =\int \frac{1+\cos x}{2} d x=\frac{1}{2}\left(\int d x+\int \cos x d x\right)= \\
& =\frac{1}{2}(x+\sin x)+C \quad[\text { II, I; (2), (8)]. }
\end{aligned}
$$

6) $\int \operatorname{tg}^{2} x d x=\int\left(\sec ^{2} x-1\right) d x=\int \frac{d x}{\cos ^{2} x}-\int d x=$ $=\operatorname{tg} x-x+C \quad[$ II; (10), (2)].

Using the main rules I, II, and integration formulas (1) - (14), find the following integrals and verify the results by differentiation: