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I calculated this integral below : $$ \int_{-\pi/2}^{\pi/2} \cos(x)\cos(nx) dx $$

And I got this result : $\dfrac{\left(\left(n-1\right)\sin\left(\frac{{\pi}n+{\pi}}{2}\right)+\left(n+1\right)\sin\left(\frac{{\pi}n-{\pi}}{2}\right)\right)}{n^2-1}$

I had a check at the end at the solution : $-\dfrac{2\cos\left(\frac{{\pi}n}{2}\right)}{n^2-1}$

So apparently there is some kind of simplification to do but I don't actually know how .

Sebastiano
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Samir
  • 247

3 Answers3

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$\sin (\frac {n\pi} 2+\frac {\pi} 2)=\cos (\frac {n \pi} 2)$ and $\sin (\frac {n\pi} 2-\frac {\pi} 2)=-\cos (\frac {n \pi} 2)$.

[$\sin (\frac {\pi} 2+\theta)=\cos (\theta)$ and $\sin (\theta -\frac {\pi} 2)=-\sin (\frac {\pi} 2-\theta)=-\cos (\theta)$]

Kavi Rama Murthy
  • 311,013
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Observe that $$\sin\left(\frac{\pi n+ \pi}{2}\right)=\sin\left(\frac{\pi n}{2}\right)\cos\left(\frac{\pi}{2}\right)+ \sin\left(\frac{\pi}{2}\right)\cos\left(\frac{\pi n}{2}\right) = \cos\left(\frac{\pi n}{2}\right)$$ $$\sin\left(\frac{\pi n- \pi}{2}\right)=\sin\left(\frac{\pi n}{2}\right)\cos\left(\frac{\pi}{2}\right)- \sin\left(\frac{\pi}{2}\right)\cos\left(\frac{\pi n}{2}\right) = -\cos\left(\frac{\pi n}{2}\right)$$ when $n\in\mathbb{N}$.

Then, \begin{equation*} \begin{split} (n-1)\sin\left(\frac{\pi n+ \pi}{2}\right) & + (n+1)\sin\left(\frac{\pi n- \pi}{2}\right) = (n-1)\cos\left(\frac{\pi n}{2}\right) - (n+1)\cos\left(\frac{\pi n}{2}\right) \\ & = -2\cos\left(\frac{\pi n}{2}\right) \end{split} \end{equation*}

user326159
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Use Prosthaphaeresis Formulas,

$$n\left(\sin\dfrac{(n+1)\pi}2+\sin\dfrac{(n-1)\pi}2\right)-\left(\sin\dfrac{(n+1)\pi}2-\sin\dfrac{(n-1)\pi}2\right)$$

$$=2n\sin\dfrac{n\pi}2\cos\dfrac\pi2-2\cos\dfrac{n\pi}2\sin\dfrac\pi2$$

Alternatively,

like Evaluate the integral $\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}\,\mathrm dx$.,

as $\cos x\cdot\cos(nx)$ is an even function using Werner Formulas,

$$\int_{-a}^a\cos x\cdot\cos(nx)\ dx=2\int_0^a\dfrac{\cos(n-1)x+\cos(n+1)x}2\ dx$$

lab bhattacharjee
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