1023. Using series, calculate the approximate values of the following integrals with an accuracy of 0.0001:

$$
I_{1}=\int_{0}^{\frac{1}{3}} \frac{d t}{\sqrt{1+t^{2}}} ; \quad I_{2}=\int_{0}^{1} \cos \sqrt{x} d x ; \quad I_{3}=\int_{1}^{1.5} \frac{1}{v} \operatorname{arctg} \frac{v}{4} d v
$$
Solution. 1) We expand the integrand into a binomial series (B), setting $x=t^{4}, m=-\frac{1}{2}$:

$$
\frac{1}{\sqrt{1+t^{4}}}=\left(1+t^{4}\right)^{-\frac{1}{2}}=1-\frac{1}{2} t^{4}+\frac{1 \cdot 3}{2 \cdot 4} t^{8}-\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} t^{12}+\ldots
$$

This series converges to the binomial $\left(1+t^{4}\right)^{-\frac{1}{2}}$ for $|t|<1$.

Integrating from 0 to $\frac{1}{2}$, we find

$$
\begin{aligned}
& l_{1}=\int_{0}^{\frac{1}{2}} \frac{d t}{\sqrt{1+t^{4}}}=t-\frac{1}{2} \frac{t^{5}}{5}+\frac{1 \cdot 3}{2 \cdot 4} \frac{t^{9}}{9}-\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \frac{t^{13}}{13}+\left.\ldots\right|_{a} ^{\frac{1}{2}}= \\
= & \frac{1}{2}-\frac{1}{2 \cdot 5 \cdot 2^{5}}+\frac{1 \cdot 3}{2 \cdot 4 \cdot 9 \cdot 2^{9}}-\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6 \cdot 13 \cdot 2^{13}}+\ldots
\end{aligned}
$$

We compute the first few terms of the resulting alternating convergent series (with one extra digit): $a_{1}=0.50000 ; a_{2} \approx-0.00313 ; a_{3} \approx 0.00008$.

According to the property of an alternating convergent series (§ 2), to compute the integral with an accuracy of 0.0001, it is sufficient to take the sum of the first two terms of the series $I_{1} \approx a_{1}+a_{2} \approx 0.4969$.

The error of this approximate value is less than the absolute value of the first discarded term of the series, i.e., less than $a_{3} \approx 0.00008$.

2) Using the Maclaurin series for $\cos x$, replacing $x$ with $\sqrt{x}$, we have

$$
\cos \sqrt{x}=1-\frac{x}{2!}+\frac{x^{2}}{4!}-\frac{x^{3}}{6!}+\frac{x^{1}}{8!}-\ldots(x \geqslant 0)
$$

Integrating over the specified limits, we get

$$
\begin{gathered}
I_{2}=\int_{0}^{1} \cos \sqrt{x} d x=x-\frac{x^{2}}{2!2}+\frac{x^{3}}{4!3}-\frac{x^{4}}{6!4}+\frac{x^{5}}{8!5}-\left.\ldots\right|_{0} ^{1}= \\
=1-\frac{1}{2!2}+\frac{1}{4!3}-\frac{1}{6!4}+\frac{1}{8!5}-\ldots
\end{gathered}
$$

The fifth term of this alternating convergent series is less than 0.0001. Therefore, to compute the desired approximate value of the integral, it is sufficient to take the sum of the first four terms of the series:

$$
I_{2} \approx 1-\frac{1}{4}+\frac{1}{72}-\frac{1}{2880} \approx 0.7635
$$

3) Using the Maclaurin series for $\operatorname{arctg} x$, for $x=\frac{v}{4}$, we get

$$
\operatorname{arctg} \frac{v}{4}=\frac{v}{4}-\frac{v^{3}}{4^{3} \cdot 3}+\frac{v^{6}}{4^{5} \cdot 5}-\frac{v^{2}}{4^{7} \cdot 7}+\ldots(|v| \leqslant 4)
$$

Dividing both sides of the equation by $v$ and integrating, we find

$$
\begin{gathered}
I_{3}=\int_{1}^{1.5} \frac{1}{v} \operatorname{arctg} \frac{v}{4} d v=\frac{v}{4}-\frac{v^{3}}{4^{3} \cdot 3^{2}}+\frac{v^{5.5}}{4^{5} \cdot 5^{2}}-\frac{v^{2}}{4^{7} \cdot 7^{2}}+\left.\ldots\right|_{1} ^{1.5}= \\
\quad=\frac{1.5-1}{4}-\frac{1.5^{3}-1}{4^{3} \cdot 3^{2}}+\frac{1.5^{5}-1}{4^{5} \cdot 5^{2}}-\frac{1.5^{2}-1}{4^{7} \cdot 7^{2}}+\ldots \approx \\
\approx 0.12500-0.00412+0.00026-0.00002+\ldots
\end{gathered}
$$

The obtained result represents an alternating convergent series. Taking the sum of its first three terms, we get the approximate value of the integral with the given accuracy: $I_{3} \approx 0.1211$, since the absolute value of the fourth term is less than 0.0001.