7.1 Integration By Parts/43: Difference between revisions
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<math> | <math> | ||
\begin{align} | \begin{align} | ||
\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + | |||
\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx \\[2ex] | |||
\end{align} | \end{align} | ||
</math> | </math> | ||
\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx | \int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx \\[2ex] | ||
&= -\frac{1}{2}\cos(x)\sin(x) + \frac{1}{2}x + c \\[2ex] | &= -\frac{1}{2}\cos(x)\sin(x) + \frac{1}{2}x + c \\[2ex] | ||
&= -\frac{1}{2} \cdot 2 \cdot \frac{1}{2}\sin(x)\cos(x) + \frac{x}{2} + c \\[2ex] | &= -\frac{1}{2} \cdot 2 \cdot \frac{1}{2}\sin(x)\cos(x) + \frac{x}{2} + c \\[2ex] | ||
&= -\frac{1}{4}\sin(2x) + \frac{x}{2} + c \\[2ex] | &= -\frac{1}{4}\sin(2x) + \frac{x}{2} + c \\[2ex] | ||
Revision as of 00:23, 29 November 2022
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int \sin ^{2}(x)dx&=-{\frac {1}{2}}\cos(x)\cdot \sin ^{2-1}x+{\frac {2-1}{2}}\int \sin ^{0}(x)dx\\[2ex]\end{aligned}}}
\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx \\[2ex]
&= -\frac{1}{2}\cos(x)\sin(x) + \frac{1}{2}x + c \\[2ex]
&= -\frac{1}{2} \cdot 2 \cdot \frac{1}{2}\sin(x)\cos(x) + \frac{x}{2} + c \\[2ex]
&= -\frac{1}{4}\sin(2x) + \frac{x}{2} + c \\[2ex]