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How to transform the integral

$$\int _{0}^{\pi }\sin ^{2}\left( \psi \right) \sin \left( m\psi \right) d\psi $$

to

$$\int _{0}^{\pi }\left( \dfrac {1} {2}-\dfrac {1} {2}\cos 2\psi \right) \sin m\psi d\psi $$

What is the general method you need to solve trig questions like this. How are do you know which identities to use and which ones should you always have memorised to derive this.

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    $\begingroup$ $\sin^2 x = \dfrac{1}{2}(1-\cos 2x)$ should be one of many in your toolbox. $\endgroup$
    – David P
    May 14, 2017 at 7:21
  • $\begingroup$ @DavidP okay that's fine when you just say it like that for someone who is familiar with this. However which ones should i always have in my toolbox, and assuming that i didn't know that how can i derive it. $\endgroup$
    – Tutorz
    May 14, 2017 at 7:28

2 Answers 2

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Some other useful tools for basic trig. integrals:

$\sin(2x)=2 \sin (x)\cos (x)$
$\cos(2x)=1-\cos^2(x)=1-2\sin^2(x)$
and the most elementary one: $1=\sin^2(x)+\cos^2(x)$,

Trigonometric products to sums and identities for higher-exponent trigonometric functions are also handy, but harder to memorize. These can be found from math-tables.

Also good rules to remember when integrating trig. functions:
$$\frac{d}{dx}(\sin^n(x))=n\cos(x)\sin^{n-1}(x)$$ (or generally) $$\int f'f^n=\frac{f^{n+1}}{n+1}+C,(n \neq-1)$$
$$\int \frac{f'}{f}=\ln|f|+C$$

So in many cases the trick is to derive the equation (with trigonometric identities) to a form where these rules can be applied.

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The general approach

Considering trigonometric integrals there are a myriad of different techniques to solve them. Sometimes integration by parts is the best option, other times a clever substitution or a trigonometric identity saves the day. Sadly, the only way to know which one to use is to experiment and solve as many problems as possible.

My toolbox

In my toolbox I only store three basic identities when it comes to integrating sine and cosine.

$$ \begin{align*} 1 & = (\sin x)^2 + (\cos x)^2 \\ \cos(A+B) & = \cos(A)\cos(B) - \sin(A) \sin(B) \\ \sin(A+B) & = \cos(A) \sin(B) + \cos(B) \sin(A) \end{align*} $$

Everything else I can derive from these pretty simple formulas. Of course as mentioned earlier I know how to use integration by parts, and use substitutions as well.

The problem at hand

Let's look at your problem and use a fairly clever trick. I start with defining the following pair of integrals

$$ I = \int_0^{\pi} (\sin \psi)^2 \sin (m\psi)\, \mathrm{d}\psi \quad \text{and} \quad J = \int_0^{\pi} (\cos \psi)^2 \sin (m\psi)\,\mathrm{d}\psi $$

Notice how simple we can solve the following integral

$$ I + J = \int_0^\pi (\sin^2 \psi + \cos^2\psi) \sin (m\psi)\, \mathrm{d}\psi = \begin{cases} \cfrac{1 - \cos m\pi}{m} & \text{if} \qquad m \neq 0 \\ 0 & \text{if} \qquad m = 0 \end{cases} $$

Similarly, the integral $J - I$ can be evaluated fairly easy with the use of $\cos^2\psi - \sin^2\psi = \cos 2\psi$ and

$$ \cos ax \sin bx = \frac{\sin(a x + b x) - \sin(a x - b x)}{2}\,. $$

This can be derived by taking the difference between $\sin(A+B)$ and $\sin(A-B)$ and then using the addition formula for sine, written at the top. Thus,

$$ \begin{align*} J - I & = \int_0^\pi (\cos^2\psi - \sin^2\psi )\sin (m\psi) \,\mathrm{d}\psi \\ & = \int_0^\pi \cos (2\psi) \sin (m\psi)\, \mathrm{d}\psi \\ & = \frac{1}{2}\int_0^\pi \sin (2 + m)\psi - \sin(2 - m)\psi \,\mathrm{d}\psi \\ & = \begin{cases} \cfrac{m}{m^2 - 4}(1 - \cos m\pi) & \text{if} \qquad m^2 \neq 4 \\ 0 & \text{if} \qquad m^2 = 4 \end{cases} \end{align*} $$

Assuming $m \neq 0$ and $m^2 \neq 2$ we then have the following system of equations

$$ \begin{align*} J - I & = \cfrac{1 - \cos m\pi}{m} \\ J + I & = \cfrac{m}{m^2 - 4}(1 - \cos m\pi) \end{align*} $$

Solving this set of equations yields

$$ I = \frac{2}{m} \frac{\cos(m\pi) - 1}{m^2 - 4} $$

Doing the same in the cases $m^2 = 4$ and $m = 0$, just give $I = 0$, thus

$$ \int_0^\pi (\sin \psi)^2 \sin(m\psi) \,\mathrm{d}\psi = \begin{cases} \cfrac{2}{m} \cfrac{\cos(m\pi) - 1}{m^2 - 4} & \text{if} \qquad m\neq -2,0,2 \\ 0 & \text{if} \qquad m = -2, 0, 2 \end{cases} $$

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