EXAMPLE 2. Calculate the definite integral

$$
\int_{0}^{\arccos (1 / \sqrt{6})} \frac{3 \tan^{2} x-1}{\tan^{2} x+5} d x
$$
Solution.

1. Since the integrand has the form $R(\operatorname{tg} x)$ and the function $\operatorname{tg} x$ is defined on $[0, \arccos (1 / \sqrt{6})]$, we make the substitution $\operatorname{tg} x=t$.

Substituting into the integrand,

$$
\operatorname{tg} x=t, \quad d x=\frac{d t}{1+t^{2}}
$$

we obtain a rational function of $t$:

$$
\frac{3 \operatorname{tg}^{2} x-1}{\operatorname{tg}^{2} x+5} d x=\frac{3 t^{2}-1}{t^{2}+5} \frac{1}{1+t^{2}} d t
$$

2. We find the new limits of integration:

$$
\operatorname{tg}(0)=0, \quad \operatorname{tg}\left(\arccos \frac{1}{\sqrt{6}}\right)=\sqrt{5}
$$

3. We apply the formula for the change of variable in a definite integral:

$$
\int_{0}^{\arccos (1 / \sqrt{6})} \frac{3 \operatorname{tg}^{2} x-1}{\operatorname{tg}^{2} x+5} d x=\int_{0}^{\sqrt{5}} \frac{3 t^{2}-1}{t^{2}+5} \cdot \frac{1}{1+t^{2}} d t
$$

4. We compute the antiderivative of the rational function of $t$ and apply the Newton-Leibniz formula:

$$
\begin{aligned}
& \int_{0}^{\sqrt{5}} \frac{3 t^{2}-1}{t^{2}+5} \cdot \frac{1}{1+t^{2}} d t=-\left.\operatorname{arctg} t\right|_{0} ^{\sqrt{5}}+\left.\frac{4}{\sqrt{5}} \operatorname{arctg} \frac{t}{\sqrt{5}}\right|_{0} ^{\sqrt{5}}=\frac{\pi}{\sqrt{5}}-\operatorname{arctg} \sqrt{5} \\
& \text { Answer. } \quad \int_{0}^{\arccos (1 / \sqrt{6})} \frac{3 \operatorname{tg}^{2} x-1}{\operatorname{tg}^{2} x+5} d x=\frac{\pi}{\sqrt{5}}-\operatorname{arctg} \sqrt{5} .
\end{aligned}
$$

Conditions of the Problems. Compute the definite integrals.

1. $\int_{0}^{\pi / 2} \frac{1}{5+4 \sin x+3 \cos x} d x$.
2. $\int_{0}^{\pi / 2} \frac{1}{3+5 \sin x+3 \cos x} d x$.
3. $\int_{\pi / 6}^{\pi / 2} \frac{\cos ^{3} x}{\sin ^{2} x+\sin x} d x$
4. $\int_{\pi / 3}^{\pi / 2} \frac{\cos x}{3+\cos x} d x$.
5. $\int_{\pi / 3}^{\pi / 2} \frac{1}{\sin x(1+\cos x)} d x$.
6. $\int_{\pi / 4}^{\pi / 2} \frac{\cos ^{2}}{\sin ^{4} x} d x$
7. $\int^{\pi / 3} \frac{1+\operatorname{tg} x}{\sin 2 x} d x$.
8. $\int_{0}^{\pi / 4} \frac{1}{4 \cos ^{2} x-2 \sin 2 x+\sin ^{2} x} d x$.
9. $\int_{0}^{\pi / 4} \frac{1}{5+\cos ^{2} x} d x$.
10. $\int_{0}^{\pi / 4} \frac{2 \sin x+3 \cos x}{\sin ^{2} x \cos x+9 \cos ^{3} x} d x$.

Answers. 1. $\frac{1}{6} \cdot 2 \cdot \frac{1}{5} \ln \frac{8}{3} \cdot 3 \cdot \ln 2-\frac{1}{2} \cdot 4 . \operatorname{arctg} \frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2} \operatorname{arctg} \frac{\sqrt{6}}{6}$.
5. $\frac{1}{2}+\frac{1}{4} \ln 3$.
6. $\frac{1}{3} \cdot \quad 7 \cdot \frac{1}{2}(\sqrt{3}-1+\ln \sqrt{3})$.
8. $\frac{1}{2}$.
9. $\frac{1}{\sqrt{30}} \operatorname{arctg} \sqrt{\frac{5}{6}}$.

10. $\ln \frac{10}{9}+\operatorname{arctg} \frac{1}{3}$.

## 8.4. Integration of Expressions

$\sin ^{2 m} x \cos ^{2 n} x$

PROBLEM STATEMENT. Compute the definite integral

$$
\int_{a}^{b} \sin ^{2 m} x \cos ^{2 n} x d x
$$

PLAN OF SOLUTION. We apply the formulas for reducing the degree $\sin ^{2} x=\frac{1}{2}(1-\cos 2 x), \quad \cos ^{2} x=\frac{1}{2}(1+\cos 2 x), \quad \sin x \cos x=\frac{1}{2} \sin 2 x$ until we arrive at tabular integrals or integrals that can be reduced to tabular ones in a known manner.

REMARK. It is useful to keep in mind that

$$
\int_{0}^{2 \pi} \sin k x d x=0, \quad \int_{0}^{2 \pi} \cos k x d x=0 \quad(k \in \mathbb{N})
$$