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Question: How do you find the integral of \({\sin ^3}\left[ x \right]dx\)?...

How do you find the integral of sin⁡3[x]dx{\sin ^3}\left[ x \right]dx?

Explanation

Solution

To find the integral of sin⁡3[x]dx{\sin ^3}\left[ x \right]dx, we need to use some relations. First of all, we can write sin⁡3[x]dx{\sin ^3}\left[ x \right]dx as sin⁡x⋅sin⁡2x\sin x \cdot {\sin ^2}x. Now, we have an identity, sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1. So, therefore sin⁡2x=1−cos⁡2x{\sin ^2}x = 1 - {\cos ^2}x. Substitute this value of sin in the obtained equation. Now, simplify the integral and you can easily integrate the function.

Complete step by step solution:
In this question, we have to find the integral of sin⁡3[x]dx{\sin ^3}\left[ x \right]dx.
Let the integral of sin⁡3[x]dx{\sin ^3}\left[ x \right]dx be
⇒I=∫sin⁡3xdx\Rightarrow I = \int {{{\sin }^3}xdx}- - - - - - - - - - - - - (1)
Now, we can write sin⁡3xdx{\sin ^3}xdx as sin⁡x⋅sin⁡2x\sin x \cdot {\sin ^2}x. Therefore, equation (1) becomes
⇒I=∫sin⁡x⋅sin⁡2xdx\Rightarrow I = \int {\sin x \cdot {{\sin }^2}xdx}- - - - - - - - - - - (2)
Now, we know the trigonometric identity that is, sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1.
Therefore, sin⁡2x=1−cos⁡2x{\sin ^2}x = 1 - {\cos ^2}x. Substituting the value of sin⁡2x{\sin ^2}x in equation (2), we get
⇒I=∫sin⁡x⋅(1−cos⁡2x)dx\Rightarrow I = \int {\sin x \cdot \left( {1 - {{\cos }^2}x} \right)dx}- - - - - - - - - - - (3)
Now, opening the bracket we get
⇒I=∫sin⁡x−sin⁡xcos⁡2xdx\Rightarrow I = \int {\sin x - \sin x{{\cos }^2}xdx}- - - - - - - - (4)
Now, separating the terms, we get
⇒I=∫sin⁡xdx−∫sin⁡xcos⁡2xdx\Rightarrow I = \int {\sin xdx - \int {\sin x{{\cos }^2}xdx} }
⇒I=∫sin⁡xdx+∫−sin⁡xcos⁡2xdx\Rightarrow I = \int {\sin xdx + \int { - \sin x{{\cos }^2}xdx} }- - - - - - - - - (5)
Now, we know that ∫sin⁡xdx=−cos⁡x\int {\sin xdx = - \cos x} .
Now, for ∫sin⁡xcos⁡2xdx\int {\sin x{{\cos }^2}xdx} , we have a result
∫f(x)n⋅f′(x)dx=f(x)n+1n+1\int {f{{\left( x \right)}^n} \cdot f'\left( x \right)dx = \dfrac{{f{{\left( x \right)}^{n + 1}}}}{{n + 1}}}
Here, f(x)=cos⁡xf\left( x \right) = \cos x and f′(x)=−sin⁡xf'\left( x \right) = - \sin x and n=2n = 2.
Therefore, ∫sin⁡xcos⁡2xdx=cos⁡2+1x2+1=cos⁡3x3\int {\sin x{{\cos }^2}xdx} = \dfrac{{{{\cos }^{2 + 1}}x}}{{2 + 1}} = \dfrac{{{{\cos }^3}x}}{3}.
Therefore, equation (5) becomes
⇒I=−cos⁡x+cos⁡3x3+c\Rightarrow I = - \cos x + \dfrac{{{{\cos }^3}x}}{3} + c
Where, c is the integration constant.
Hence, the integral of sin⁡3[x]dx{\sin ^3}\left[ x \right]dx is −cos⁡x+cos⁡3x3+c - \cos x + \dfrac{{{{\cos }^3}x}}{3} + c.

Note:
Note that there is no direct formula for finding the integral of trigonometric functions with power greater than 1. So, we always need to use trigonometric relations and formulas to bring the function in simple terms so that we can integrate it easily.