How do I evaluate the indefinite integral $\int sin (3 x) \cdot sin (6 x) d x$ $intsin(3x)*sin(6x)dx$ ?

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How do I evaluate the indefinite integral $\int sin (3 x) \cdot sin (6 x) d x$ $intsin(3x)*sin(6x)dx$ ?

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Answer 1 · 2014-08-13T04:47:46

It can have two solutions

$= \frac{sin 3 x}{6} - \frac{sin 9 x}{18} + c$ $=(sin3x)/6-(sin9x)/18+c$ , where $c$ $c$ is a constant OR

$= \frac{2}{9} {sin}^{3} (3 x) + c$ $=2/9sin^3(3x)+c$ , where $c$ $c$ is a constant

Explanation

$= \int sin (3 x) \cdot sin (6 x) d x$ $=intsin(3x)*sin(6x)dx$

From trigonometric identities,

$sin A sin B = \frac{1}{2} (cos (A - B) - cos (A + B))$ $sinAsinB=1/2(cos(A-B)-cos(A+B))$

Similarly, for the given problem,

$sin (3 x) \cdot sin (6 x) = \frac{1}{2} (cos (- 3 x) - cos 9 x)$ $sin(3x)*sin(6x)=1/2(cos(-3x)-cos9x)$

As, $cos (- A) = cos (A)$ $cos(-A)=cos(A)$

$sin (3 x) \cdot sin (6 x) = \frac{1}{2} (cos 3 x - cos 9 x)$ $sin(3x)*sin(6x)=1/2(cos3x-cos9x)$

integrating both sides,

$\int sin (3 x) \cdot sin (6 x) d x = \int \frac{1}{2} (cos 3 x - cos 9 x) d x$ $intsin(3x)*sin(6x)dx=int1/2(cos3x-cos9x)dx$

$= \frac{1}{2} \int (cos 3 x) d x - \frac{1}{2} \int (cos 9 x) d x$ $=1/2int(cos3x)dx-1/2int(cos9x)dx$

$= \frac{1}{2} \frac{sin 3 x}{3} - \frac{1}{2} \frac{sin 9 x}{9} + c$ $=1/2(sin3x)/3-1/2(sin9x)/9+c$ , where $c$ $c$ is a constant

$= \frac{sin 3 x}{6} - \frac{sin 9 x}{18} + c$ $=(sin3x)/6-(sin9x)/18+c$ , where $c$ $c$ is a constant

Another Method :

$= \int sin (3 x) \cdot sin (6 x) d x$ $=intsin(3x)*sin(6x)dx$

$= \int sin (3 x) \cdot 2 sin (3 x) cos (3 x) d x$ $=intsin(3x)*2sin(3x)cos(3x)dx$

$= \int 2 {sin}^{2} (3 x) cos (3 x) d x$ $=int2sin^2(3x)cos(3x)dx$

let's assume $sin (3 x) = t$ $sin(3x)=t$ , then $3 cos (3 x) d x = d t$ $3cos(3x)dx=dt$

therefore,

$= 2 \int \frac{t^{2}}{3} \cdot d t$ $=2intt^2/3*dt$

$= \frac{2}{9} \cdot t^{3} + c$ $=2/9*t^3+c$ , where $c$ $c$ is a constant

$= \frac{2}{9} {sin}^{3} (3 x) + c$ $=2/9sin^3(3x)+c$ , where $c$ $c$ is a constant

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