## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{1} \frac{x^{4} \cdot d x}{\left(2-x^{2}\right)^{3 / 2}}
$$
## Solution

$$
\int_{0}^{1} \frac{x^{4} \cdot d x}{\left(2-x^{2}\right)^{3 / 2}}=
$$

Substitution:

$$
\begin{aligned}
& x=\sqrt{2} \sin t \Rightarrow d x=\sqrt{2} \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{\sqrt{2}}=0 \\
& x=1 \Rightarrow t=\arcsin \frac{1}{\sqrt{2}}=\frac{\pi}{4}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{0}^{\pi / 4} \frac{4 \sin ^{4} t \cdot \sqrt{2} \cos t}{\left(2-2 \sin ^{2} t\right)^{3 / 2}} d t=\int_{0}^{\pi / 4} \frac{4 \sqrt{2} \sin ^{4} t \cdot \cos t}{2 \sqrt{2}\left(1-\sin ^{2} t\right)^{3 / 2}} d t=\int_{0}^{\pi / 4} \frac{2 \sin ^{4} t \cdot \cos t}{\cos ^{3} t} d t= \\
& =2 \cdot \int_{0}^{\pi / 4} \frac{\sin ^{4} t}{\cos ^{2} t} d t=2 \cdot \int_{0}^{\pi / 4} \frac{\left(1-\cos ^{2} t\right)^{2}}{\cos ^{2} t} d t=2 \cdot \int_{0}^{\pi / 4} \frac{1-2 \cos ^{2} t+\cos ^{4} t}{\cos ^{2} t} d t= \\
& =2 \cdot \int_{0}^{\pi / 4}\left(\frac{1}{\cos ^{2} t}-2+\frac{1+\cos 2 t}{2}\right) d t=\int_{0}^{\pi / 4}\left(\frac{2}{\cos ^{2} t}-3+\cos 2 t\right) d t= \\
& =\left.\left(2 \operatorname{tg} t-3 t+\frac{1}{2} \sin 2 t\right)\right|_{0} ^{\pi / 4}=\left(2 \operatorname{tg} \frac{\pi}{4}-\frac{3 \pi}{4}+\frac{1}{2} \sin \frac{\pi}{2}\right)-\left(2 \operatorname{tg} 0-3 \cdot 0+\frac{1}{2} \sin 0\right)= \\
& =\left(2 \cdot 1-\frac{3 \pi}{4}+\frac{1}{2} \cdot 1\right)-\left(2 \cdot 0-0+\frac{1}{2} \cdot 0\right)=\frac{5}{2}-\frac{3 \pi}{4}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\� \%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5 $\% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+12-9$ »

Categories: Kuznetsov's Problem Book Integrals Problem 12 | Integrals

- Last modified: 11:25, 5 July 2009.
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## Problem Kuznetsov Integrals 12-10

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