Example 1.1. Find the integrals:
a) $\int \sin (7 x+3) d x$,
b) $\int \frac{d x}{x+5}$.
c) $\int(3 x+5)^{99} d x$,
d) $\int e^{x^{3}} x^{2} d x$
e) $\int \sin ^{3} x \cos x d x$.
Solution. a) Here the substitution $t=7 x+3$ is made, as in this case the indefinite integral is reduced to a tabular integral of sine

$$
\begin{aligned}
t=7 x+3 \Rightarrow d t & =d(7 x+3)=(7 x+3)^{\prime} d x=7 d x, \text { i.e. } \\
d t & =7 d x \Rightarrow d x=\frac{1}{7} d t
\end{aligned}
$$

Taking into account the chosen substitution, we get

$$
\int \sin (7 x+3) d x=\frac{1}{7} \int \sin t d t=-\frac{1}{7} \cos t+C=-\frac{1}{7} \cos (7 x+3)+C
$$

b) In this example, the substitution $t=x+5$ is natural, then

$$
d t=d(x+5)=(x+5)^{\prime} d x=d x \Rightarrow d t=d x
$$

As a result

$$
\int \frac{d x}{(x+5)}=\int \frac{d t}{t}=\ln |t|+C=\ln |x+5|+C
$$

c) This indefinite integral is easily found by the substitution $t=3 x+5$. Obviously, $d t=3 d x$ and

$$
\int(3 x+5)^{99} d x=\frac{1}{3} \int t^{99} d t=\frac{1}{3} \frac{t^{100}}{100}+C=\frac{(3 x+5)^{100}}{300}+C
$$

d) In this case, the substitution $x=t^{3}$ leads to the goal. Then

$$
d t=d x^{3}=\left(x^{3}\right)^{\prime} d x=3 x^{2} d x \Rightarrow x^{2} d x=\frac{1}{3} d t
$$

Taking into account these results

$$
\int e^{x^{3}} x^{2} d x=\frac{1}{3} \int e^{t} d t=\frac{1}{3} e^{t}+C=\frac{1}{3} e^{x^{3}}+C
$$

This integral can also be found directly by bringing $x^{2}$ under the differential sign, considering that by property (1.3) $d \int f(x) d x=f(x) d x$. The expression $x^{2} d x$ can be rewritten as

$$
x^{2} d x=d \int x^{2} d x=d \frac{x^{3}}{3}=\frac{1}{3} d x^{3}
$$

Then the integral in question can be rewritten as follows: $\int e^{x^{3}} x^{2} d x=\frac{1}{3} \int e^{x^{3}} d x^{3}$. If we now denote $u=x^{3}$, we get in the end

$$
\int e^{x^{3}} x^{2} d x=\frac{1}{3} \int e^{x^{3}} d x^{3}=\frac{1}{3} \int e^{u} d u=\frac{1}{3} e^{u}+C=\frac{1}{3} e^{x^{3}}+C
$$

Remark. Formula (1.9) reflects the property of invariance of the integration formula: integration formulas retain their form when substituting any differentiable function of the independent variable for the independent variable.

e) It is easy to notice that $\cos x d x=d \sin x$. Then

$$
\int \sin ^{3} x \cos x d x=\int \sin ^{3} x d \sin x
$$

If we now denote $\sin x=t$, we get

$$
\int \sin ^{3} x \cos x d x=\int t^{3} d t=\frac{t^{4}}{4}+C=\frac{\sin ^{4} x}{4}+C
$$

In the general case, the transformation of the integrand expression to a new variable involves two aspects:

1) transforming the integrand function in terms of the new variable
2) expressing the differential of the old variable in terms of the differential of the new variable.

This last step is carried out by formally finding the differential of the right and left sides of the equations
$x=\varphi(t)$ or $t=\varphi^{-1}(x)$, connecting the old and new variables:

$$
d x=d \varphi(t)=\varphi^{\prime}(t) d t \text { or } d t=d \varphi^{-1}(x)=\left(\varphi^{-1}(x)\right)^{\prime} d x
$$