How do you find the integral of ${(sin x)}^{3} d x$ $(sinx)^3 dx$ ?

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How do you find the integral of ${(sin x)}^{3} d x$ $(sinx)^3 dx$ ?

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Answer 1 · 2015-09-25T02:20:09

Refer to explanation

Explanation:

We know that

$sin 3 x = 3 sin x - 4 {(sin x)}^{3} \Rightarrow 4 {(sin x)}^{3} = 3 sin x - sin 3 x \Rightarrow {(sin x)}^{3} = \frac{3}{4} \cdot sin x - \frac{1}{4} sin 3 x$ $sin 3x = 3sinx - 4 (sin x)^3=>4(sinx)^3=3sinx-sin3x=> (sinx)^3=3/4*sinx-1/4sin3x$

Hence we have that

$\int {(sin x)}^{3} d x = \int (\frac{3}{4} \cdot sin x - \frac{1}{4} \cdot sin (3 x)) d x = - \frac{3}{4} cos x + \frac{1}{12} cos 3 x + c$ $int (sinx)^3 dx=int (3/4*sinx-1/4*sin(3x))dx=-3/4cosx+1/12cos3x+c$

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How do you find the integral of ${(sin x)}^{3} d x$ $(sinx)^3 dx$ ?

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