7.1 Integration By Parts/45: Difference between revisions

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<math>


\begin{align}




&\int_{0}^{\frac{\pi}{2}} sin^5(x) dx \\[2ex]
<math>


&= \frac{5-1}{5} \int_{0}^{\frac{\pi}{2}} sin^3(x) dx \\[2ex]
\begin{align}
& u= cos(x)          \\[2ex]
&  du= -sin(x) \\[2ex]
&  dx= \frac{du}{-sin(x)} \\[2ex]
& \int sin^3(x) dx \\[2ex]
&= \int (1-cos^2(x)) sin(x) dx \\[2ex]
&= \int (1-u^2)sin(x) \cdot \frac{du}{-sin(x)} \\[2ex]
&= -\int 1-u^2 du \\[2ex]
&= \frac{4}{5} [-u + \frac{1}{3}^3(x)] \\[2ex]
&= \frac{4}{5} [-cos(x) + \frac{1}{3} cos^3(x)] \\[2ex]
&= \frac{4}{5} [-cos(\frac{\pi}{2}) + \frac{1}{3} cos^3 (\frac{\pi}{2}) - (-cos(0) + \frac{1}{3} cos^3 (0)) \\[2ex]
&= \frac{4}{5} [(-1)(0) + \frac{1}{3} (0)^3 - ((-1)(1) +\frac{1}{3} cos^3(1) \\[2ex]
 
&= \frac{4}{5} [-(-1+ \frac{1}{3}(1)] \\[2ex]
&= \frac{4}{5}[\frac{2}{3}] \\[2ex]
&= \frac{8}{15} \\[2ex]


& \int_{0}^{\frac{\pi}{2}} sin^{2n+1} (x) dx \\[2ex]
&= \frac{2n+1-1}{2n+1} \int_{0}^{\frac{\pi}{2}} sin^{2n+1-2} (x) dx \\[2ex]
&= \frac{2n}{2n+1} \int_{0}^{\frac{\pi}{2}} sin^{2n-1} (x) dx \\[2ex]




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</math>
</math>


<math>
<math>
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\begin{align}
\begin{align}


& \int_{0}^{\frac{\pi}{2}} sin^{2n+1} (x) dx \\[2ex]
\int_{0}^{\frac{\pi}{2}} sin^5(x) dx &= \frac{5-1}{5} \int_{0}^{\frac{\pi}{2}} sin^3(x) dx \\[2ex]
&= \frac{2n+1-1}{2n+1} \int_{0}^{\frac{\pi}{2}} sin^{2n+1-2} (x) dx \\[2ex]
&= \frac{4}{5}\int_{0}^{\frac{\pi}{2}} (1-cos^2(x)) sin(x) dx \\[2ex]
&= \frac{2n}{2n+1} \int_{0}^{\frac{\pi}{2}} sin^{2n-1} (x) dx \\[2ex]
&= -\int_{1}^{0} (1-u^2)\cdot du \\[2ex]
&= \frac{4}{5} \left[u - \frac{u^3}{3}\right]\bigg|_{0}^{1} \\[2ex]




& u= cos(x)          \\[2ex]
&  du= -sin(x) \\[2ex]
&  dx= \frac{du}{-sin(x)} \\[2ex]
& \int sin^3(x) dx \\[2ex]
&= \frac{4}{5} [1-\frac{1}{3}] \\[2ex]
&= \frac{8}{15}





Latest revision as of 00:30, 30 November 2022

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} & \int_{0}^{\frac{\pi}{2}} sin^n(x) dx = \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2} (x) dx \\[2ex] &= -\frac{1}{n} cos(x) sin^{n-1} (x) + \frac{n-1}{n} \bigg|_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex] &= -\frac{1}{n} cos(\frac{\pi}{2}) sin^{n-1} (\frac{\pi}{2})+ \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex] &= -\frac{1}{n} (0) (1) + \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex] &= \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} sin^{n-2}(x) dx \\[2ex] \end{align} }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} & \int_{0}^{\frac{\pi}{2}} sin^3(x) dx \\[2ex] &= \frac{3-1}{3} \int_{0}^{\frac{\pi}{2}} sin(x) dx \\[2ex] &= \frac{2}{3} \int_{0}^{\frac{\pi}{2}} sin(x) dx \\[2ex] &= \frac{2}{3} [-cos(x) \bigg|_{0}^{\frac{\pi}{2}} \\[2ex] &= \frac{2}{3} [-cos(\frac{\pi}{2}) -(-cos(0))] \\[2ex] &= \frac{2}{3} [-1(0) -(-1)] \\[2ex] &= \frac{2}{3}[1] \\[2ex] &= \frac{2}{3} \\[2ex] \end{align} }



Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} & \int_{0}^{\frac{\pi}{2}} sin^{2n+1} (x) dx \\[2ex] &= \frac{2n+1-1}{2n+1} \int_{0}^{\frac{\pi}{2}} sin^{2n+1-2} (x) dx \\[2ex] &= \frac{2n}{2n+1} \int_{0}^{\frac{\pi}{2}} sin^{2n-1} (x) dx \\[2ex] \end{align} }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} \int_{0}^{\frac{\pi}{2}} sin^5(x) dx &= \frac{5-1}{5} \int_{0}^{\frac{\pi}{2}} sin^3(x) dx \\[2ex] &= \frac{4}{5}\int_{0}^{\frac{\pi}{2}} (1-cos^2(x)) sin(x) dx \\[2ex] &= -\int_{1}^{0} (1-u^2)\cdot du \\[2ex] &= \frac{4}{5} \left[u - \frac{u^3}{3}\right]\bigg|_{0}^{1} \\[2ex] & u= cos(x) \\[2ex] & du= -sin(x) \\[2ex] & dx= \frac{du}{-sin(x)} \\[2ex] & \int sin^3(x) dx \\[2ex] &= \frac{4}{5} [1-\frac{1}{3}] \\[2ex] &= \frac{8}{15} \end{align} }