Definite Integral of sin^4x from 0 to pi/4

In this tutorial we shall derive definite integral of trigonometric function \sin^{4}x from limits 0 to Pi/4.

The integration of the form

\begin{gathered} I = \int\limits_0^{\frac{\pi }{4}} {{{\sin }^4}xdx} \\ \Rightarrow I = \int\limits_0^{\frac{\pi }{4}} {{{\left( {{{\sin }^2}x} \right)}^2}dx} \\ \end{gathered}

Using half angle formula from trigonometry {\sin ^2}x = \frac{{1 - \cos 2x}}{2}, we have

\begin{gathered} I = \int\limits_0^{\frac{\pi }{4}} {{{\left( {\frac{{1 - \cos 2x}}{2}} \right)}^2}dx} \\ \Rightarrow I = \frac{1}{4}\int\limits_0^{\frac{\pi }{4}} {{{\left( {1 - \cos 2x} \right)}^2}dx} \\ \Rightarrow I = \frac{1}{4}\int\limits_0^{\frac{\pi }{4}} {\left( {1 - 2\cos 2x + {{\cos }^2}2x} \right)dx} \\ \Rightarrow I = \frac{1}{4}\int\limits_0^{\frac{\pi }{4}} {dx} - \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\cos 2xdx} + \frac{1}{4}\int\limits_0^{\frac{\pi }{4}} {{{\cos }^2}2xdx} \\ \end{gathered}

Again using half angle formula from trigonometry {\cos ^2}2x = \frac{{1 + \cos 4x}}{2}, we have

\begin{gathered} I = \frac{1}{4}\int\limits_0^{\frac{\pi }{4}} {dx} - \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\cos 2xdx} + \frac{1}{4}\int\limits_0^{\frac{\pi }{4}} {\left( {\frac{{1 + \cos 4x}}{2}} \right)dx} \\ \Rightarrow I = \frac{1}{4}\int\limits_0^{\frac{\pi }{4}} {dx} - \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\cos 2xdx} + \frac{1}{8}\int\limits_0^{\frac{\pi }{4}} {dx + \frac{1}{8}\int\limits_0^{\frac{\pi }{4}} {\cos 4x} dx} \\ \end{gathered}

First we evaluate this integration by using integral formula \int {\cos kxdx = \frac{{\sin kx}}{k}} , then we using the basic rule of definite integral \int\limits_a^b {f\left( x \right)dx = \left| {F\left( x \right)} \right|_a^b} = \left[ {F\left( b \right) - F\left( a \right)} \right], we have

\begin{gathered} \int\limits_0^{\frac{\pi }{4}} {{{\sin }^4}xdx} = \left| x \right|_0^{\frac{\pi }{4}} - \frac{1}{2}\left| {\frac{{\sin 2x}}{2}} \right|_0^{\frac{\pi }{4}} + \frac{1}{8}\left| x \right|_0^{\frac{\pi }{4}} + \frac{1}{8}\left| {\frac{{\sin 4x}}{4}} \right|_0^{\frac{\pi }{4}} \\ \Rightarrow \int\limits_0^{\frac{\pi }{4}} {{{\sin }^4}xdx} = \frac{1}{4}\left[ {\frac{\pi }{4} - 0} \right] - \frac{1}{4}\left[ {\sin 2\frac{\pi }{4} - \sin 0} \right] + \frac{1}{8}\left[ {\frac{\pi }{4} - 0} \right] + \frac{1}{{32}}\left[ {\sin 4\frac{\pi }{4} - \sin 0} \right] \\ \Rightarrow \int\limits_0^{\frac{\pi }{4}} {{{\sin }^4}xdx} = \frac{\pi }{{16}} - \frac{1}{4}\left[ {\sin \frac{\pi }{2} - 0} \right] + \frac{1}{{32}} + \frac{1}{{32}}\left[ {\sin \pi - 0} \right] \\ \Rightarrow \int\limits_0^{\frac{\pi }{4}} {{{\sin }^4}xdx} = \frac{\pi }{{16}} - \frac{1}{4}\left( 1 \right) + \frac{\pi }{{32}} + \frac{1}{{32}}\left( 0 \right) \\ \Rightarrow \int\limits_0^{\frac{\pi }{4}} {{{\sin }^4}xdx} = \frac{{3\pi - 8}}{{32}} \\ \end{gathered}