7.1 Integration By Parts/24: Difference between revisions

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<math>
\begin{align}


&\int_{0}^{\pi}\underbrace{x^3\cos(x)}_{
\begin{aligned}
u&=x^3 \quad \quad dv=\cos(x) \\
dv&=3x^2 \quad \quad v=\sin(x)
\end{aligned}}
\,dx =x^3\sin(x)-\int_{0}^{\pi} \underbrace{3x^2\sin(x)}_{
\begin{aligned}
u&=3x^2 \quad \quad dv=\sin(x) \\
du&=6x \quad \quad v=-\cos(x)
\end{aligned}}
\,dx= x^3\sin(x)-[3x^2-\cos(x)-\int_{0}^{\pi}-6x\cos(x)\,dx]\\
=&x^3\sin(x)-3x^2\cos(x)-\int_{0}^{\pi}6x\cos(x)
\end{align}
</math>

Revision as of 19:11, 1 December 2022