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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.

  • Is it $sin^2(x)or sin(x^2)$ make parentheses clear – Archis Welankar Apr 02 '16 at 15:49
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    Integration by parts. – André Nicolas Apr 02 '16 at 15:54
  • Doesn't that eventually yield $\int \sin^2 x dx = \int \cos^2 x dx$ so that the reasoning is circular (no pun intended)? – jim Apr 02 '16 at 16:00
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    @jim : There's a way out of your problem. See my answer below. $\qquad$ – Michael Hardy Apr 02 '16 at 16:09
  • @ArchisWelankar : Please: write $\sin^2 x$ rather than $sin^2 x$. It is coded as \sin^2 x. The backslash not only prevents italicization but also results in proper spacing in things like $a\sin b$ and $a\sin(b)$ (notice less space to the right in the latter than in the former). $\qquad$ – Michael Hardy Apr 02 '16 at 16:14
  • Thanks for formatting suggestion. As for the answer of the question I like varied responses –  Apr 02 '16 at 16:21
  • There is actually much to be said for writing $(\sin x)^2$ rather than $\sin^2 x$ (since $\sin^2 x$ arguably ought to mean $\sin(\sin(x))$, but convention doesn't agree with that). My comment about MathJax code was intended to mean that the backslash should be used. $\qquad$ – Michael Hardy Apr 02 '16 at 16:35

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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.

Michael Hardy
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