Question

How do you integrate \[\sin 3x\cos 3xdx\]?

Answer 1

Hint: In solving the given question, we have to integrate the given function, we first consider one part of the question as a variable, here it will be \[u = \sin 3x\], now derive the function using the identity,\[\dfrac{d}{{dx}}\sin x = \cos x\] and we will derive the function again then we will get the second part of the function. Now we will get a function which will be easily integrated and that will be our required result.

Complete step-by-step answer:
Given function is \[\sin 3x\cos 3xdx\],
We first consider \[u = {e^{\sin x}}\],
Now differentiating on both sides we get,
\[ \Rightarrow \]\[\dfrac{d}{{dx}}u = \dfrac{d}{{dx}}\sin 3x\],
Now using the identity \[\dfrac{d}{{dx}}\sin x = \cos x\] and if the expression has two functions then we have to again derive the second function also, we get,
\[ \Rightarrow \]\[\dfrac{{du}}{{dx}} = \cos 3x\dfrac{d}{{dx}}3x\],
Now using the identity \[\dfrac{d}{{dx}}ax = a\], we get,
\[ \Rightarrow \]\[\dfrac{{du}}{{dx}} = \cos 3x \cdot 3\],
Now taking \[dx\] to the right hand side of the expression we get,
\[ \Rightarrow \]\[du = 3\cos 3xdx\],
Now substituting the values in the given function becomes,
\[\sin 3x\cos 3x = u \cdot \dfrac{1}{3}\],
Now applying integration on both sides we get,
\[ \Rightarrow \int {\sin 3x\cos 3x = } \int {u \cdot \dfrac{1}{3}du} \],
Now using the integral identity\[\int {xdx = \dfrac{{{x^2}}}{2} + c} \], we get,
\[ \Rightarrow \int {\sin 3x\cos 3xdx} = \dfrac{1}{3}\left( {\dfrac{{{u^2}}}{2}} \right) + c\],
Now substituting u value i.e.,\[u = \sin 3x\] as we first considered, in the above expression we get,
\[ \Rightarrow \int {\sin 3x\cos 3xdx} = \dfrac{1}{3}\left( {\dfrac{{{{\sin }^2}3x}}{2}} \right) + c\]
Now simplifying we get,
\[ \Rightarrow \int {\sin 3x\cos 3xdx} = \dfrac{{{{\sin }^2}3x}}{6} + c\],
So, the integral of \[\sin 3x\cos 3xdx\] is equal to \[\dfrac{{{{\sin }^2}3x}}{6} + c\].

\[\therefore \]The integral value of \[\sin 3x\cos 3xdx\] will be equal to \[\dfrac{{{{\sin }^2}3x}}{6} + c\].

Note:
In this type of question we use both derivation and integration formulas as both are related to each other so students should not get confused where to use the formulas and which one to use as there are many formulas in both derivation and integration. Some of the important formulas that are used are given below:
\[\dfrac{d}{{dx}}x = 1\],
\[\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}\],
\[\dfrac{d}{{dx}}uv = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\],
\[\dfrac{d}{{dx}}{e^x} = {e^x}\],
\[\dfrac{d}{{dx}}\dfrac{u}{v} = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}\],
\[\dfrac{d}{{dx}}\ln x = \dfrac{1}{x}\],
\[\dfrac{d}{{dx}}\sin x = \cos x\],
\[\dfrac{d}{{dx}}\cos x = - \sin x\],
\[\int {dx = x + c} \],
\[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + c\],
\[\int {\dfrac{1}{x}dx} = \ln x + c\],
\[\int {xdx} = \dfrac{{{x^2}}}{2} + c\].