Question

Integrate sin (2pix+30 degree) dx

Answer 1

we have to integrate $\int{sin(2\pi x+30^{\circ})}\,dx$

first of all, we have to resolve sin(2πx + 30°)

we know, sin(A + B) = sinA.cosB + cosA.sinB

so, sin(2πx + 30°) = sin(2πx).cos30° + sin30°.cos(2πx)

= sin(2πx). (√3/2) + (1/2). cos(2πx)

now, $\int{sin(2\pi x+30^{\circ})}\,dx=\frac{\sqrt{3}}{2}\int{sin(2\pi x)}\,dx+\frac{1}{2}\int{cos(2\pi x)}\,dx$

$=\frac{\sqrt{3}}{2}\left[\frac{-cos(2\pi x)}{2\pi}\right]+\frac{1}{2}\left[\frac{sin(2\pi x)}{2\pi}\right]$

$=-\frac{\sqrt{3}}{4\pi}cos(2\pi x)+\frac{1}{4\pi}sin(2\pi x)$

$=\frac{1}{2\pi}\left(-\frac{\sqrt{3}}{2}cos(2\pi x)+\frac{1}{2}sin(2\pi x)\right)$

$=\frac{1}{2\pi}\left(cos(150^{\circ}).cos(2\pi x)+sin(150^{\circ}).sin(2\pi x)\right)$

$=\frac{1}{2\pi}cos(2\pi x - 150^{\circ})$

[ we know, cosA.cosB + sinA.sinB = cos(A - B) ]

hence, answer is$=\frac{1}{2\pi}cos(2\pi x - 150^{\circ})$

Answer 2

Explanation:

answer is -1/2π cos(2πx + 30°) + C