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