Problem archive/8/8.1.tex

% Calculus, Integration, TrigonometricSubstitution, Substitution

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\begin{document}
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\section*{Problem Statement}

Oblicz 
\[
\int \left( \frac{x^4}{1+x^6} \right)^2 \, dx
\]

\bigskip

\noindent\textbf{Solution:}

$$\int \left( \frac{x^4}{1+x^6} \right)^2 \, dx = \int \left( \frac{x^8}{\left(1+\left(x^3\right)^2\right)^2} \right) \, dx$$

Niech $x^3 = \tan\theta \Rightarrow x^6 = \tan^2\theta$. Dodatkowo $\theta = \arctan\left(x^3\right)$.

\begin{align*}
x^3 &= \tan\theta \\
d(x^3) &= d(\tan\theta) \\
3x^2\,dx &= \sec^2\theta\,d\theta \\
dx &= \frac{\sec^2\theta\,d\theta}{3x^2} 
\end{align*}

Zatem podstawmy to do naszej pierwotnej całki.

\begin{align*}
\int \left( \frac{x^8}{\left(1+\left(x^3\right)^2\right)^2} \right) \, dx &= \int \left( \frac{x^8}{\left(1+\left(x^3\right)^2\right)^2} \right)\cdot\frac{\sec^2\theta\,d\theta}{3x^2} \\
&= \frac{1}{3}\int\frac{x^6}{\left(1+\tan^2\theta\right)^2}\sec^2\theta\,d\theta \\
&= \frac{1}{3}\int\frac{x^6}{\sec^4\theta}\sec^2\theta\,d\theta \\
&= \frac{1}{3}\int\frac{\tan^2\theta}{\sec^2\theta}\,d\theta \\
&= \frac{1}{3}\int\frac{\frac{\sin^2\theta}{\cos^2\theta}}{\frac{1}{\cos^2\theta}}\,d\theta 
\end{align*}

Dalej na następnej stronie.

\begin{align*}
&= \frac{1}{3}\int\frac{\frac{\sin^2\theta}{\cos^2\theta}}{\frac{1}{\cos^2\theta}}\,d\theta \\
&= \frac{1}{3}\int\frac{\sin^2\theta}{\cos^2\theta}\cos^2\theta\,d\theta \\
&= \frac{1}{3}\int\sin^2\theta\,d\theta \\
&= \frac{1}{3}\int\frac{1}{2}\left(1-\cos2\theta\right)\,d\theta \\
&= \frac{1}{6}\left(\theta - \frac{1}{2}\sin2\theta\right) \\
&= \frac{1}{6}\theta - \frac{1}{6}\sin\theta\cos\theta
\end{align*}

\begin{minipage}[T]{0.45\textwidth}
    \raggedright
    $\tan\theta = x^3 \Leftrightarrow \tan\theta = \frac{x^3}{1}$

    Z trójkąta po prawej mamy:
    \begin{align*}
        \sin\theta &= \frac{x^3}{\sqrt{x^6+1}} \\
        \cos\theta &= \frac{1}{\sqrt{x^6+1}} \\
    \end{align*}
\end{minipage}
\hfill
\begin{minipage}[T]{0.45\textwidth}
    \raggedright
    \begin{tikzpicture}[scale=1.3]
        % Coordinates
        \coordinate (A) at (0,0);
        \coordinate (B) at (4,0); 
        \coordinate (C) at (0,3); 

        % Triangle
        \draw[thick] (A) -- (B) -- (C) -- cycle;

        % Right angle mark
        \draw (0.3,0) -- (0.3,0.3) -- (0,0.3);

        % Labels for sides
        \node[below] at ($(A)!0.5!(B)$) {$1$}; % Base
        \node[left] at ($(A)!0.5!(C)$) {$x^3$};  % Height
        \node[above right] at ($(B)!0.5!(C)$) {$\sqrt{x^6 + 1}$}; % Hypotenuse

        % Angle theta at vertex B
        \draw[-, thick] (B) ++(135:0.5) arc[start angle=135, end angle=180, radius=0.6];
        \node at ($(B)+(150:0.85)$) {$\theta$};

    \end{tikzpicture}
\end{minipage}

Podstawiając to do tamtego wyniku otrzymujemy dalej:
\begin{align*}
    \frac{1}{6}\theta - \frac{1}{6}\sin\theta\cos\theta &= \frac{1}{6}\arctan x^3 - \frac{1}{6}\frac{x^3}{\sqrt{x^6+1}}\frac{1}{\sqrt{x^6+1}} \\
    &= \frac{1}{6}\arctan x^3 - \frac{1}{6}\frac{x^3}{x^6+1} \\
    &= \frac{1}{6}\arctan x^3 - \frac{x^3}{6x^6+6} + C, \quad C\in\mathbb{R} 
\end{align*}

\textbf{Odpowiedź:} 
$$\int \left( \frac{x^4}{1+x^6} \right)^2 \, dx = \frac{1}{6}\arctan x^3 - \frac{x^3}{6x^6+6} + C, \quad C\in\mathbb{R}$$

\end{document}