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

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}&\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{aligned}}}