7.1 Integration By Parts/43: Difference between revisions

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<math> \text{Prove with the reduction formula that} \int\sin^2(x)dx = \frac{x}{2} - \frac{sin(2x)}{4} + C </math> <br><br>
<math> \text{Prove with the reduction formula that} \int\sin^2(x)dx = \frac{x}{2} - \frac{sin(2x)}{4} + C </math> <br>
<math> \text{reduction formula:} \int\sin^ndx= -\frac{1}{n}\cos(x)\sin^{n-1}(x) + \frac{n-1}{n}\int\sin^{n-2}(x)dx </math>
<math> \text{reduction formula:} \int\sin^ndx= -\frac{1}{n}\cos(x)\sin^{n-1}(x) + \frac{n-1}{n}\int\sin^{n-2}(x)dx </math>
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Revision as of 00:26, 29 November 2022


Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\text{reduction formula:}}\int \sin ^{n}dx=-{\frac {1}{n}}\cos(x)\sin ^{n-1}(x)+{\frac {n-1}{n}}\int \sin ^{n-2}(x)dx} 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]&=-{\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]\end{aligned}}}