Problems on Techniques of Integration

We have a definite integral which depends on the integer $n=1,2,\cdots$ The fundamental formula $\sin^2(x) + \cos^2(x) = 1$ will be very useful. We have

\begin{displaymath}\begin{array}{lll}
I_{n+2} &=&\displaystyle \int_0^{\pi/2} \s...
...aystyle I_n - \int_0^{\pi/2}\sin^n(x)\cos^2(x)dx\\
\end{array}\end{displaymath}

In order to evaluate $\displaystyle \int_0^{\pi/2}\sin^n(x)\cos^2(x)dx$, we will use the integration by parts technique. Since the derivative of $\sin^k(x)$ is $k \sin^{k-1}(x) \cos(x)$, we set

\begin{displaymath}\left\{\begin{array}{lll}
u &=& \cos(x)\\
dv &=& \sin^n(x)\cos(x) dx\;.
\end{array}\right.\end{displaymath}

Then

\begin{displaymath}\left\{\begin{array}{lll}
du &=&-\sin(x)dx\\
v &=& \displaystyle \frac{1}{n+1} \sin^{n+1}(x)\;.
\end{array}\right.\end{displaymath}

Since

\begin{displaymath}\int_a^b u dv = [u v]_a^b - \int_a^b v du\;,\end{displaymath}

we get

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

Since $\sin(0) = 0$ and $\cos(\pi/2) = 0$, we get

\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}

So

\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}

This is the recurrent formula. In particular, the first integrals $I_0 = \pi/2$ and $I_1 = 1$, will give all the integrals $I_n$, for $n=1,2,\cdots$ For example, we have

\begin{displaymath}I_2 = \frac{1}{2} I_0 = \frac{\pi}{4}\;\;\;\mbox{and}\;\; I_3 = \frac{2}{3}I_1 = \frac{2}{3}\;\cdot\end{displaymath}


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