Question

How do you find the antiderivative of ${\cos ^2}3xdx$?

Answer 1

Hint: The word antiderivative means the inverse of the given function. Therefore, we must find the antiderivative of the given trigonometric expression which is also equal to finding the integral of the expression. Expand the trigonometric expression in terms of simpler expressions to easily find the integral.

Formula used: $\cos 2\theta = 2{\cos ^2}\theta - 1$
$\int {\cos axdx = \dfrac{{\sin ax}}{a}}$

Complete step-by-step solution:
The given trigonometric expression is ${\cos ^2}3xdx$
Now we must find the integral of that expression as we are asked to find the antiderivative of it.
When we tag along with the integral symbol, we rewrite the expression as,
$\Rightarrow \int {{{\cos }^2}3xdx}$
Firstly, we simplify the term ${\cos ^2}3x$
Since we know that,
$\cos 2\theta = 2{\cos ^2}\theta - 1$
We put $\theta = 3x$
Then we get the expression,
$\Rightarrow \cos 2(3x) = 2{\cos ^2}(3x) - 1$
On further evaluation we get,
$\Rightarrow \cos (6x) = 2{\cos ^2}(3x) - 1$
Now since the term we need is ${\cos ^2}(3x)$,
We rewrite the expression putting the term we need on LHS.
$\Rightarrow \dfrac{{\cos (6x) + 1}}{2}$
Now write the same with the integral.
$\Rightarrow \int {\dfrac{{\cos (6x) + 1}}{2}} dx$
Let’s split the equation for a better evaluation of the integral.
$\Rightarrow \int {\dfrac{{\cos (6x)}}{2}} dx + \int {\dfrac{1}{2}dx}$
Since the integral of $\cos x$ is $\sin x$
And $\int {\cos axdx = \dfrac{{\sin ax}}{a}}$
$\Rightarrow \dfrac{1}{2}\int {\cos (6x)dx = \dfrac{{\sin 6x}}{6}}$
Now on considering the second term,
The integral of $\int 1 dx = x$
$\Rightarrow \int {\dfrac{1}{2}dx} = \dfrac{1}{2}x$
Now on putting it all together,
$\Rightarrow \left[ {\dfrac{1}{2} \times \dfrac{{\sin (6x)}}{6}} \right] + \dfrac{1}{2}x + C$
Now simplify further.
$\Rightarrow \dfrac{1}{2}\left[ {\dfrac{{sin(6x)}}{6} + x} \right] + C$
$\Rightarrow \dfrac{{sin(6x) + 6x}}{{12}} + C$

Finally, we can now conclude that the antiderivative of,
$\therefore {\cos ^2}3xdx = \dfrac{{sin(6x) + 6x}}{{12}} + C$

Additional Information: Whenever complex equations are given to solve one must always Firstly start from the complex side and then convert all the terms into $\sin \theta$ or $\cos \theta$ . Then combine them into single fractions. Now it’s most likely to use Trigonometric identities for the transformations if there are any. Know when and where to apply the Subtraction-Addition formula.

Note: Never forget to write $dx\;$ whenever we are writing the expression under the integral. Also, after the whole solution is written, there should be a mention of the other constants which will be denoted by $C$ . On evaluation the integral and the $dx\;$ get canceled.