How do I calculate this definite integral?

Question

Can someone tell me how to calculate the given definite integral $$\int_{-\pi}^\pi \sin^3(t) \sin(nt)dt$$

Taking by-parts would be hell.

Answer 1

hint

The integrand function is even, you need to compute $$2\int_0^{\pi}\sin^3(t)\sin(nt)dt$$

use the linearisation

$$\sin^3(t)=\frac 14\Bigl(3\sin(t)-\sin(3t)\Bigr)$$

and the transformation $$\sin(a)\sin(b)=$$ $$\frac 12\Bigl(\cos(a-b)-\cos(a+b)\Bigr)$$

Answer 2

If you write $ \sin^3 (x) $ as $ \frac 1 4 \big(3 \sin(x) - \sin(3x)\big) $ (a pretty identity derived from the calculations of $\big(\cos(x) +i \sin(x)\big) ^3 $ ), you can get $\int_{-\pi} ^ \pi \sin^3(x) \sin(nx) \mathrm dx = \\ = \frac{1}{4} \left( \int_{-\pi} ^ \pi 3\sin (x) \sin(nx) \mathrm dx - \int_{-\pi} ^ \pi \sin(3x)\sin(nx) \mathrm dx \right) = \\ = \frac{\pi}{4} ( 3\delta_{ 1n} - \delta_{3n} ) $ where $\delta_{ij} $ is the Kronecker's delta, and the last identity follows from the ortogonality of the Fourier trigonometric system. I hope it helps. Sorry for the edits, I had some sign and parentheses errors.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ In the absence of any information about $\ds{n}$, I'll assume $\ds{n \in \mathbb{Z}}$. \begin{align} &\bbox[5px,#ffd]{% \int_{-\pi}^{\pi}\sin^{3}\pars{t}\sin\pars{nt}\,\dd t} = \Im\int_{-\pi}^{\pi}\sin^{3}\pars{t}\expo{\ic nt}\,\dd t \\[5mm] = &\ \Im\oint_{\verts{z}\ =\ 1}\pars{z - 1/z \over 2\ic}^{3}z^{n}\, {\dd z \over \ic z} = {1 \over 8}\,\Im\oint_{\verts{z}\ =\ 1} {\pars{z^{2} - 1}^{3} \over z^{4 - n}}\,\dd z \\[5mm] = &\ {1 \over 8}\,\Im\oint_{\verts{z}\ =\ 1} \pars{{1 \over z^{-2 - n}} - 3\,{1 \over z^{-n}} + 3\,{1 \over z^{2 - n}} - {1 \over z^{4 - n}}}\,\dd z \\[5mm] = &\ \bbx{{1 \over 4}\,\pi\pars{\delta_{n,-3} - 3\delta_{n,-1} + 3\delta_{n1} - \delta_{n3}}} \\ &\ \end{align}