Solveeit Logo

Question

Question: How do I find the indefinite integral of \(\sin (3x)\cos (3x)?\)...

How do I find the indefinite integral of sin(3x)cos(3x)?\sin (3x)\cos (3x)?

Explanation

Solution

In this question, you have to find the indefinite integral of the given trigonometric function. To find the integral, firstly use the trigonometric identity for half angle formula of sine function to simplify the given trigonometric function then integrate the simplified function with help of basic integration formulas for trigonometry.Half angle formula for sine function is given as, sin2x=2sinxcosx\sin 2x = 2\sin x\cos x.

Complete step by step answer:
In order to find the indefinite integral of the given trigonometric function sin(3x)cos(3x)\sin (3x)\cos (3x) we will first simplify the given trigonometric function with help of half angle formula for sine function as follows,
sin(3x)cos(3x)\sin (3x)\cos (3x)
Dividing and multiplying the above expression with two, we will get
12×2sin(3x)cos(3x)\dfrac{1}{2} \times 2\sin (3x)\cos (3x)
Now, we can see that the above expression is looking similar to the trigonometric identity of half angle formula for sine function that is given as sin2x=2sinxcosx\sin 2x = 2\sin x\cos x, so using this we can further simplify the expression as
12×sin(6x)\dfrac{1}{2} \times \sin (6x)
Now integrating it, we will get
12×sin(6x)dx 12×sin(6x)dx 12×(cos(6x)6+c) \int {\dfrac{1}{2} \times \sin (6x)dx} \\\ \Rightarrow \dfrac{1}{2} \times \int {\sin (6x)dx} \\\ \Rightarrow \dfrac{1}{2} \times \left( {\dfrac{{ - \cos (6x)}}{6} + c} \right) \\\
Simplifying this, we will get
cos(6x)12+12×c cos(6x)12+C \dfrac{{ - \cos (6x)}}{{12}} + \dfrac{1}{2} \times c \\\ \therefore\dfrac{{ - \cos (6x)}}{{12}} + C \\\
Therefore cos(6x)12+C\dfrac{{ - \cos (6x)}}{{12}} + C is the required indefinite integral of the given function.

Additional information:
We have written 12×c=C\dfrac{1}{2} \times c = C in the integral part because the constant we get when doing the indefinite integration is an arbitrary constant and we can consider anything to be an arbitrary constant and also algebraic operation does not affect arbitrary constants.

Note: When integrating a term consisting of a constant then you can write the constant outside the integral as we have written in this question. Also take care of writing the constant part after doing indefinite integral.