432. Find the integrals
1) $\int(3-2 x)^{2} d x$
2) $\int \sec ^{2}(m-n x) d x$
3) $\int \operatorname{tg} \varphi d \varphi$.
Solution. 1) We multiply and divide by -2, introduce the factor -2 under the integral sign according to property III, and replace $-2 dx$ with $d(3-2x)$, which is the same, to get:

$\int(3-2 x)^{7} d x=-\frac{1}{2} \int(3-2 x)^{7}(-2 d x)=-\frac{1}{2} \int(3-2 x)^{7} d(3-2 x)=$

$$
=-\frac{1}{2} \cdot \frac{(3-2 x)^{8}}{8}+C
$$

by formula 1, with $u=3-2 x, a=7$.

2) Multiplying and dividing by $-n$ and using the equality $-n dx=$ $=d(m-n x)$, we find:

$$
\begin{aligned}
& \int \sec ^{2}(m-n x) d x=-\frac{1}{n} \int \sec ^{2}(m-n x) \cdot(-n d x)= \\
= & -\frac{1}{n} \int \sec ^{2}(m-n x) d(m-n x)=-\frac{1}{n} \operatorname{tg}(m-n x)+C
\end{aligned}
$$

by formula 6, with $u=m-n x$.

3) Replacing $\operatorname{tg} \varphi$ with $\frac{\sin \varphi}{\cos \varphi}$, we get:

$$
\begin{gathered}
\int \operatorname{tg} \varphi d \varphi=\int \frac{\sin \varphi}{\cos \varphi} d \varphi=-\int \frac{-\sin \varphi}{\cos \varphi} d \varphi=-\int \frac{d \cos \varphi}{\cos \varphi}= \\
=-\ln |\cos \varphi|+C
\end{gathered}
$$

by formula 2, with $u=\cos \varphi\left(u^{\prime}=-\sin \varphi, d u=-\sin \varphi d \varphi\right)$.

It is useful to remember the verbal expression of formula 2: the integral of a fraction whose numerator is the differential of the denominator is equal to the logarithm of the absolute value of the denominator.

Find the following integrals and verify the results by differentiation:

(To save space in the answers to the problems in this chapter, the arbitrary constant $C$ is omitted.)