Other way of integration of $(\sin^2 x)$

Question

I was wondering if there is any other way of integrating $(\sin^2 x)$ without using formula of $\cos 2 x$.

Please avoid expansion series in your answers. Thanking in advance.

Answer 1

Here's one way: \begin{align} & \int (\sin x) \Big( \sin x\, dx\Big) = \overbrace{\int u\,dv = uv - \int v\,du}^\text{integration by parts} \\[10pt] = {} & (\sin x)(-\cos x) - \int (-\cos x)\Big(\cos x\,dx\Big) \\[10pt] = {} & -\sin x\cos x + \int \cos^2 x \,dx \\[10pt] = {} & -\sin x \cos x + \int (1-\sin^2 x)\,dx \\[10pt] = {} & -\sin x \cos x + \int 1\,dx - \int \sin^2 x\, dx \\[10pt] = {} & x - \sin x\cos x - \int \sin^2 x \,dx. \end{align} Thus, letting $\displaystyle I= \int \sin^2 x\,dx$, we have $$ I = x - \sin x \cos x - I. $$ Adding $I$ to both sides we get $$ 2I = x - \sin x \cos x + \text{constant} $$ (since the two $\text{“}I\text{''}$s may differ by a constant). Then divide both sides by $2$.

$$\S$$

Here's a remark on the definite integral. \begin{align} & \int_0^{\pi/2} \sin^2 x\,dx = \int_{\pi/2}^0 (\cos^2 u)\,(-du) \quad \left(\text{where } u = \frac \pi 2 - x\right) \\[10pt] = {} & \int_0^{\pi/2} (\cos^2 x)\,dx. \end{align} But $$ \int_0^{\pi/2} \sin^2 x\,dx + \int_0^{\pi/2} \cos^2 x\,dx = \int_0^{\pi/2} 1\,dx = \frac \pi 2. $$ Since the two integrals are equal to each other and their sum is $\dfrac\pi2$, each, must be equal to $\dfrac\pi4$.

This is one of the simplest ways of evaluating an integral without ever finding an antiderivative. Note that one need not find an antiderivative in order to evaluate $\displaystyle\int_0^{\pi/2} 1\,dx$ since the area under the graph is merely the area of a rectangle.