How is this proven using integration by parts? $\displaystyle \int{tan^n{x}}\,dx = \frac{tan^{n-1}{x}}{n-1} - \int{tan^{n-2}{x}}\,dx$ n does not equal 1
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Originally Posted by Mike9182 How is this proven using integration by parts? $\displaystyle \int{tan^n{x}}\,dx = \frac{tan^{n-1}{x}}{n-1} - \int{tan^{n-2}{x}}\,dx$ Don't need parts. $\displaystyle {\tan ^n (x)dx = \int {\tan ^{n - 2} (x)\left( {\sec ^2 (x) - 1} \right)dx} = \dfrac{{\tan ^{n - 1} (x)}} {{n - 1}} - \int {\tan ^{n - 2} (x)dx} } $