Practicing Success
Practicing Success
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$I_n=\int_0^{π/4}tan^nx\,dx$, then $\underset{n→∞}{\lim}n[I_n+I_{n+2}]$ equals: |
$\frac{1}{2}$ 1 ∞ Zero |
1 |
$I_{n+1}=\int_0^{π/4}tan^{n+1}θ\,dθ=\int_0^{π/4}tan^{n-1}θ\,sec^2θ\,dθ-\int_0^{π/4}tan^{n-1}θ\,dθ$ $I_{n+1}+I_{n-1}=\frac{tanθ}{n}|\int_0^{π/4}⇒n(I_{n+1}+I_{n-1})=1$ |
