7.1 Integration By Parts/50: Difference between revisions

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Prove
Prove
<math>
<math>
\int_{}^{} \sec^{n}x = \frac{\tan(x)}{n-1}  
\int_{}^{} \sec^{n}x = \frac{\tan(x) \cdot \sec^{n-2}(x)}{n-1} + \frac{n-2}{n-1} \int_{}^{} \sec^{n-2}(x)dx
</math>
</math>


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&= x \ln(x)^{n} - \int_{}^{} \left(n \ln(x)^{n-1} \right)dx \\[2ex]
&= x \ln(x)^{n} - \int_{}^{} \left(n \ln(x)^{n-1} \right)dx \\[2ex]
&= x \ln(x)^{n} - n \int_{}^{} \left(\ln(x)^{n-1} \right)dx \\[2ex]
&= x \ln(x)^{n} - n \int_{}^{} \left(\ln(x)^{n-1} \right)dx \\[2ex]
\int_{}^{} \sec^{n}x = \frac{\tanx \sec^{n-2}x}{n-1} + \frac{n-2}{n-1} \int_{}^{} \sec^{n-2}xdx

Revision as of 18:26, 29 November 2022

Prove

<math> \begin{align}

\int_{}^{} \left(\ln(x)^{n}\right)dx &= x \ln(x)^{n} - \int_{}^{} \left((x \frac{n \ln(x)^{n-1}}{x}) \right)dx \\[2ex] &= x \ln(x)^{n} - \int_{}^{} \left(n \ln(x)^{n-1} \right)dx \\[2ex] &= x \ln(x)^{n} - n \int_{}^{} \left(\ln(x)^{n-1} \right)dx \\[2ex]