Problems on Techniques of Integration

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

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

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

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

Then

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

Then

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

which implies

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

So

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

which implies

$$I_{n+2} = \frac{n+1}{n+2} I_n\;\cdot$$

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