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Question: How do you integrate \(\int {\left( {{{\cos }^3}x} \right)} dx\)?...

How do you integrate ∫(cos⁡3x)dx\int {\left( {{{\cos }^3}x} \right)} dx?

Explanation

Solution

This is a question of exact integral and the trick here is to break the function in such a way that one part becomes the differential to the variable of the first part wherein we can solve it easily. Also note that as the function is totally in cos, we need to take at least one part of it to sin form to do what we intend.

Complete step by step answer:
 cos⁡2x=1−sin⁡2x d(sin⁡x)dx=cos⁡x  \ {\cos ^2}x = 1 - {\sin ^2}x \\\ \dfrac{{d\left( {\sin x} \right)}}{{dx}} = \cos x \\\ \
In this question, we have to work with cos⁡3x{\cos ^3}x. We can approach this in two ways. Either we can go for the cube formulas of sin and cos which might also give the answer of this question or we can go for a very simple tactic which is to break the cube into square times one.
cos⁡3x=cos⁡2x⋅cos⁡x{\cos ^3}x = {\cos ^2}x \cdot \cos x
The benefit of doing this is that now we can use the equation, cos⁡2x=1−sin⁡2x{\cos ^2}x = 1 - {\sin ^2}x and observe how if we differentiate sin⁡x\sin x then it becomes cos⁡x\cos x which we already have in the equation. As such, our integral equation becomes
 ∫(cos⁡3x)dx=∫(cos⁡2x)cos⁡xdx =∫(1−sin⁡2x)cos⁡dx  \ \int {\left( {{{\cos }^3}x} \right)} dx = \int {\left( {{{\cos }^2}x} \right)\cos x} dx \\\ = \int {\left( {1 - {{\sin }^2}x} \right)} \cos dx \\\ \
Now, assuming sin⁡x=u\sin x = u, dudx=cos⁡x\dfrac{{du}}{{dx}} = \cos x or du=cos⁡xdxdu = \cos xdx
Putting all this into the equation, we get
 ∫(cos⁡3x)dx=∫(1−sin⁡2x)cos⁡dx =∫(1−u2)du =u−u33+C  \ \int {\left( {{{\cos }^3}x} \right)} dx = \int {\left( {1 - {{\sin }^2}x} \right)} \cos dx \\\ = \int {\left( {1 - {u^2}} \right)} du \\\ = u - \dfrac{{{u^3}}}{3} + C \\\ \
But we know that u=sin⁡xu = \sin x and as such the equation here becomes,
 ∫(cos⁡3x)dx=u−u33+C =sin⁡x−sin⁡3x3+C  \ \int {\left( {{{\cos }^3}x} \right)} dx = u - \dfrac{{{u^3}}}{3} + C \\\ = \sin x - \dfrac{{{{\sin }^3}x}}{3} + C \\\ \

Note: In case of sin and cos, whenever we face a cube, the first choice is to always move back to square times one form as that will make the equation much simpler to integrate. If that is not possible, try to move to forms where sin and cos are in 2x or 3x. If that is also not possible, then only try to move to complex integrations.