1258. Compute the integrals:
1) $\int_{0}^{\frac{\pi}{2}} \sin ^{3} x d x$;
2) $\int_{0}^{\ln 2} \sqrt{e^{x}-1} d x$
3) $\int_{-a}^{a} x^{2} \sqrt{a^{2}-x^{2}} d x$;
4) $\int_{1}^{2} \frac{\sqrt{x^{2}-1}}{x} d x$.
Solution. 1) The given integral can be easily reduced to an integral of the form

$$
\int f(\cos x) \sin x d x
$$

therefore, we apply the substitution $\cos x=t, \sin x d x=-d t$.

Determine the new interval of integration. If $x=0$, then $\cos 0=t$ and $t=1$; if $x=\frac{\pi}{2}$, then $\cos \frac{\pi}{2}=t$ and $t=0$, hence,

$$
\begin{gathered}
\int_{0}^{\frac{\pi}{2}} \sin ^{3} x d x=\int_{0}^{\frac{\pi}{2}}\left(1-\cos ^{2} x\right) \sin x d x=-\int_{1}^{0}\left(1-t^{2}\right) d t= \\
=\int_{0}^{1}\left(1-t^{2}\right) d t=\left.\left(t-\frac{t^{3}}{3}\right)\right|_{0} ^{1}=1-\frac{1}{3}=\frac{2}{3}
\end{gathered}
$$

2) Apply the substitution $t=\sqrt{e^{x}-1}, t^{2}=e^{x}-1, \quad 2 t d t=$ $=e^{x} d x, d x=\frac{2 t}{t^{2}+1} d t$.

If $x=0$, then $t=0$; if $x=\ln 2$, then $t=\sqrt{e^{\ln 2}-1}=$ $=\sqrt{2-1}=1$, hence,

$$
\begin{gathered}
\int_{0}^{\ln 2} \sqrt{e^{x}-1} d x=\int_{0}^{1} t \frac{2 t}{t^{2}+1} d t=2 \int_{0}^{1}\left(1-\frac{1}{t^{2}+1}\right) d t= \\
=\left.2(t-\operatorname{arctg} t)\right|_{0} ^{1}=2(1-\operatorname{arctg} 1)=2-\frac{\pi}{2}
\end{gathered}
$$

3) Apply the substitution $x=a \sin t, d x=a \cos t d t$.

If $x=-a$, then $-a=a \sin t, \sin t=-1$ and $t=-\frac{\pi}{2}$, if $x=a$, then $a=a \sin t, \sin t=1$ and $t=\frac{\pi}{2}$, hence,

$$
\begin{gathered}
\int_{-a}^{a} x^{2} \sqrt{a^{2}-x^{2}} d x=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} a^{2} \sin ^{2} t \sqrt{a^{2}-a^{2} \sin ^{2} t} a \cos t d t= \\
=a^{4} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^{2} t \cos ^{2} t d t=\frac{a^{4}}{4} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^{2} 2 t d t=\frac{a^{4}}{8} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(1-\cos 4 t) d t= \\
=\left.\frac{a^{4}}{8}\left(t-\frac{\sin 4 t}{4}\right)\right|_{-\frac{\pi}{2}} ^{\frac{\pi}{2}}=\frac{a^{4}}{8}\left[\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right]=\frac{1}{8} \pi a^{4} .
\end{gathered}
$$

4) Apply the substitution $\sqrt{x^{2}-1}=t, x^{2}-1=t^{2}, 2 x d x=$ $=2 t d t$ or $x d x=t d t$.

If $x=1$, then $t=0$, if $x=2$, then $t=\sqrt{3}$, hence,

$$
\begin{aligned}
& \int_{1}^{2} \frac{\sqrt{x^{2}-1}}{x} d x=\int_{1}^{2} \frac{x \sqrt{x^{2}-1}}{x^{2}} d x=\int_{0}^{\sqrt{3}} \frac{t^{2} d t}{t^{2}+1}= \\
= & \int_{0}^{\sqrt{3}}\left(1-\frac{1}{t^{2}+1}\right) d t=\left.(t-\operatorname{arctg} t)\right|_{0} ^{\sqrt{3}}=\sqrt{3}-\frac{\pi}{3}
\end{aligned}
$$

Using appropriate substitutions, compute the integrals: