Question

Find an antiderivative (or integral) of the given function by the method of inspection $\cos 3x$

Answer 1

Hint: Let $f\left( x \right)$ be a continuous function then if there exist a function $F\left( x \right)$ which satisfy the condition $f\left( x \right)=\dfrac{d}{dx}\left( F\left( x \right) \right)$ then $F\left( x \right)$ is said to be the anti-derivative of the function $f\left( x \right)$. Use this result and also formula of derivative of trigonometric function and chain rule of derivative. That is If $y=g\left( u \right)$ be a function where $u$ is a function of $x$ then $\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}$.

Complete step-by-step answer:
Here we have to find the anti- derivative (or integral) of the given function by the method of inspection $\cos 3x$.
$\cos 3x$ is trigonometric function and trigonometric function is continuous in its domain.
Let $f\left( x \right)=\cos 3x$. So $f\left( x \right)$ is a continuous function.
Now a function $F\left( x \right)$ is said to be anti-derivative of the function $f\left( x \right)$ if $f\left( x \right)=\dfrac{d}{dx}\left( F\left( x \right) \right)$.
That is we can say that the anti-derivative of the function $f\left( x \right)$ is a function of $x$ whose derivative is $f\left( x \right)$.
Therefore the anti-derivative of $f\left( x \right)=\cos 3x$ is a function of $x$ whose derivative is $\cos 3x$.
We know the $\dfrac{d}{dx}\left( \sin x \right)=\cos x$.
And chain rule of derivative is: If $y=g\left( u \right)$ be a function where $u$ is a function of $x$ then $\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}$.
Using the above formulas of derivative we get $\dfrac{d}{dx}\left( \sin 3x \right)=\cos 3x.\dfrac{d}{dx}\left( 3x \right)$.
We know that $\dfrac{d}{dx}\left( 3x \right)=3\dfrac{d}{dx}\left( x \right)=3$
Therefore $\dfrac{d}{dx}\left( \sin 3x \right)=3\cos 3x$.
Multiply both side of above equation by $\dfrac{1}{3}$ we get $\dfrac{1}{3}.\dfrac{d}{dx}\left( \sin 3x \right)=\cos 3x$.
Also we know that if $a$ be a non-zero constant then $a.\dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{d}{dx}\left( a.f\left( x \right) \right)$.
Then using this result we get $\dfrac{d}{dx}\left( \dfrac{1}{3}.\sin 3x \right)=\cos 3x$.
Rearranging the above equation we get $\cos 3x=\dfrac{d}{dx}\left( \dfrac{1}{3}.\sin 3x \right)$.
Hence we can see that the derivative of $\dfrac{1}{3}.\sin 3x$ is $\cos 3x$.
Therefore the antiderivative (or integral) of $\cos 3x$ is $\dfrac{1}{3}.\sin 3x$.
This is the required solution.

Note: Since anti-derivative and integral both are the same thing so an alternate method to solve this problem is: finding the integral of the given function. Since integration of $\cos \left( ax \right)$ is $\dfrac{1}{a}\sin \left( ax \right)$ where $a$ is non-zero constant. Using this formula of integration we get integration of $\cos 3x$ as $\dfrac{1}{3}\sin 3x$. Also while solving this question students must take care of the basic formulas of trigonometric functions.