7.1 Integration By Parts/24: Difference between revisions

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

Revision as of 20:09, 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)-{\bigg [}3x^{2}-\cos(x)-\int _{0}^{\pi }-6x\cos(x)\,dx{\bigg ]}\\=&x^{3}\sin(x)-3x^{2}\cos(x)-\int _{0}^{\pi }\underbrace {6x\cos(x)} _{\begin{aligned}u&=6x\quad \quad dv=cos(x)\\du&=6\quad \quad v=sin(x)\end{aligned}}=x^{3}\sin(x)+3x^{2}\cos(x)-{\bigg [}6x\sin(x)-\int _{0}^{\pi }6\sin(x)\,dx{\bigg ]}\end{aligned}}}