Problems on Techniques of Integration

Use the formula $\sin^2(x) + \cos^2(x) = 1$ to get

\begin{displaymath}I_{n+2} = \int_0^{\pi/2} \sin^n(x)dx - \int_0^{\pi/2}\sin^n(x)\cos^2(x)dx = I_n - \int_0^{\pi/2}\sin^n(x)\cos^2(x)dx\;.\end{displaymath}

For the integral $\displaystyle \int_0^{\pi/2}\sin^n(x)\cos^2(x)dx$, we use the integration by parts technique. Set

u &=& \cos(x)\\
dv &=& \sin^n(x)\cos(x) dx\;.


du &=&-\sin(x)dx\\
v &=& \displaystyle \frac{1}{n+1} \sin^{n+1}(x)\;.


\begin{displaymath}\int_0^{\pi/2}\sin^n(x)\cos^2(x)dx = \left[\frac{1}{n+1} \sin...
..._0^{\pi/2} + \int_0^{\pi/2}\frac{1}{n+1} \sin^{n+1}(x)\sin(x)dx\end{displaymath}

which implies

\begin{displaymath}\int_0^{\pi/2}\sin^n(x)\cos^2(x)dx = \frac{1}{n+1} \int_0^{\pi/2}\sin^{n+2}(x)dx = \frac{1}{n+1} I_{n+2}\;.\end{displaymath}


\begin{displaymath}I_{n+2} = I_n - \int_0^{\pi/2}\sin^n(x)\cos^2(x)dx = I_n - \frac{1}{n+1} I_{n+2}\end{displaymath}

which implies

\begin{displaymath}I_{n+2} = \frac{n+1}{n+2} I_n\;\cdot\end{displaymath}

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