7.1 Integration By Parts/43: Difference between revisions
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<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 </math> | <math> | ||
\begin{align} | |||
\int\sin^2(x)dx &= - \frac{1}{2}\cos(x) \cdot \sin^{2-1}x + \frac{2-1}{2} \int\sin^{0}(x)dx </math> \\[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 | |||
Revision as of 23:54, 28 November 2022
Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \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