Answer to Question #218958 in Calculus for judas

Question

Given I_n=$\int$ tanⁿxdx show that I_n=(tan^n-1x/n-1)-I_n-2

Answer 1

$I{n}=\int tan ^n(x)dx=\int tan^{n-2}(x)tan^2(x)dx\newline =\int tan^{n-2}(x)(sec^2(x)-1)dx\newline=\int tan^{n-2}(x)sec^2(x)dx-\int tan^{n-2}(x)dx\newline let\ u=tan(x)\newline\frac{du}{dx}=sec^2(x)dx\newline du=sec^2(x)dx\newline \int tan^{n-2}(x)sec^2(x)dx=\int u^{n-2}du=\frac{u^{n-1}}{n-1}+C\newline therefore\int tan^n(x)dx=\frac{tan^{n-1}}{n+1}(x)-\int tan^{n-2}(x)dx\newline hence\ I_{n}=\frac{tan^{n-1}(x)}{n-1}-I_{n-2}$