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How would you evaluate the integral

$$\int_0 ^{\pi} \left(\frac{\pi}{2} - x\right)\sin\left(\frac{3x}{2}\right)\csc\left(\frac{x}{2}\right) dx$$

The answer from Wolfram is $0$.

Would you use a substitution or do it by parts? Would making the substitution $u=\dfrac x2$ help?

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  • $\begingroup$ You may wish to look up the triple angle formula for $\sin$. $\endgroup$ – peterwhy Nov 30 '14 at 17:03
  • $\begingroup$ WolframAlpha evaluates it to $\large\color{#c00000}{4}$. $\endgroup$ – Felix Marin Dec 30 '14 at 21:00
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Here are the steps $$ \int_0 ^{\pi} \left(\frac{\pi}{2} - x\right)\sin\left(\frac{3x}{2}\right)\csc\left(\frac{x}{2}\right) dx $$ $$= \int_0 ^{\pi} \left(\frac{\pi}{2} - x\right)\left(\frac{\sin\left(\frac{3x}{2}\right)}{\sin\left(\frac{x}{2}\right)}\right) dx $$ $$= \int_0 ^{\pi} \left(\frac{\pi}{2} - x\right)\left(\frac{3\sin\left(\frac{x}{2}\right)-4\sin^3\left(\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)}\right) dx $$ $$= \int_0 ^{\pi} \left(\frac{\pi}{2} - x\right)\left(3-4\sin^2\left(\frac{x}{2}\right)\right) dx $$ $$= \int_0 ^{\pi} \left(\frac{\pi}{2} - x\right)\left(3-4\left(\frac{1-\cos(x)}{2}\right)\right) dx $$ $$= \int_0 ^{\pi} \left(\frac{\pi}{2} - x\right)\left(3-2\left(1-\cos x\right)\right) dx $$ $$= \int_0 ^{\pi} \left(\frac{\pi}{2} - x\right)\left(1+2\cos x\right) dx $$ $$= \frac{\pi}{2}\int_0 ^{\pi}dx+\pi \int_0 ^{\pi}\cos x\ dx -\int_0^{\pi}x\ dx -2\int_0^{\pi}x\cos x\ dx $$ $$= \frac{\pi}{2}\bigg[x\bigg]_0 ^{\pi}+\pi \bigg[\sin x\bigg]_0 ^{\pi}-\bigg[\frac{x^2}{2}\bigg]_0^{\pi} -2\bigg[x\sin x+\cos x\bigg]_0^{\pi} $$ $$= \frac{\pi^2}{2}+\pi\sin \pi-\frac{\pi^2}{2}-2\bigg(\pi\sin \pi+\cos \pi-1\bigg) $$ $$= \pi\sin \pi-2\bigg(\pi\sin \pi+\cos \pi-1\bigg) $$ $$= 0-2\bigg(0-1-1\bigg) = -2\bigg(-2\bigg) =4$$

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Using the triple angle formula for sine and the power reduction formula for the square of cosine, we have:

$$\begin{align} \sin{\left(3\theta\right)} &=3\sin{\theta}-4\sin^3{\theta}\\ &=\sin{\theta}\left(3-4\sin^2{\theta}\right)\\ &=\sin{\theta}\left(4\cos^2{\theta}-1\right)\\ &=\sin{\theta}\left[4\left(\frac{1+\cos{\left(2\theta\right)}}{2}\right)-1\right]\\ &=\sin{\theta}\left[1+2\cos{\left(2\theta\right)}\right]\\ \implies \sin{\left(3\theta\right)}\csc{\left(\theta\right)}&=1+2\cos{\left(2\theta\right)};~\theta\in\mathbb{R}\land\frac{\theta}{\pi}\notin\mathbb{Z}.\\ \end{align}$$

Letting $\theta=\frac{x}{2}$ then gives us the identity,

$$\sin{\left(\frac{3x}{2}\right)}\csc{\left(\frac{x}{2}\right)}=1+2\cos{x};~x\in(2k\pi,2(k+1)\pi),k\in\mathbb{Z}.$$

Substituting $u=\frac{\pi}{2}-x$ transforms the interval of integration into a symmetric interval about the origin, and so the integral of the odd component of the integrand automatically becomes zero:

$$\begin{align} \mathcal{I} &=\int_{0}^{\pi}\left(\frac{\pi}{2}-x\right)\sin{\left(\frac{3x}{2}\right)}\csc{\left(\frac{x}{2}\right)}\,\mathrm{d}x\\ &=\int_{0}^{\pi}\left(\frac{\pi}{2}-x\right)\left(1+2\cos{x}\right)\,\mathrm{d}x\\ &=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}u\left(1+2\sin{u}\right)\,\mathrm{d}u\\ &=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}u\,\mathrm{d}u+2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}u\sin{u}\,\mathrm{d}u\\ &=0+4\int_{0}^{\frac{\pi}{2}}u\sin{u}\,\mathrm{d}u=4\int_{0}^{\frac{\pi}{2}}u\sin{u}\,\mathrm{d}u.\\ \end{align}$$

At this stage, we know that the integral $\mathcal{I}$ cannot be zero because $u\sin{u}$ is positive on $u\in(0,\frac{\pi}{2}]$. Finally, evaluating integrals of products powers and trig functions is a common application of integration by parts, of which the last integral above is an easy example:

$$\begin{align} \mathcal{I} &=4\int_{0}^{\frac{\pi}{2}}u\sin{u}\,\mathrm{d}u\\ &=4\left[\left[-u\cos{u}\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}(-1)\cos{u}\,\mathrm{d}u\right]\\ &=4\left[-\frac{\pi}{2}\cdot0+0\cdot1+\int_{0}^{\frac{\pi}{2}}\cos{u}\,\mathrm{d}u\right]\\ &=4\left[0+\left[\sin{u}\right]_{0}^{\frac{\pi}{2}}\right]\\ &=4\left[1-0\right]=4.~\blacksquare\\ \end{align}$$

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Hint

Note that: $$\sin (3x/2) \csc (x/2) = 1+2\cos x$$ Then by integration by parts you easily get the result, ($4$ not $0$ btw).

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