Question

Find the integral of the given trigonometric function \[{\sin ^2}(3x)\] ?

Answer 1

Hint: Integral of trigonometric functions are already defined but for basic identity, if you find any changes in the angle given the you have to first expand or convert the given identity into basic then you can further integrate the function, for example integral of “sinx” is “-cosx+c”, “c” is used as integral constant.

Formulae Used:
\[\Rightarrow \cos 2x = 1 - 2{\sin ^2}x \\
\Rightarrow {\sin ^2}x = \dfrac{{1 - \cos 2x}}{2} \]

Complete step by step solution:
The given question is \[{\sin ^2}(3x)\]
After implying integral sign we get:
\[\int {{{\sin }^2}\left( {3x} \right)dx} \]
Here we have to use substitution method to solve the question, in substitution method we have to assume a variable say “u” such that “3x” is equal to the variable to “u”
Now on further solving we get,
\[\dfrac{{du}}{{dx}} = 3\dfrac{{d(x)}}{{dx}} \\
\Rightarrow \dfrac{{du}}{{dx}} = 3(1) \\
\Rightarrow dx = \dfrac{1}{3}du \\ \]
Now substituting these values we get:
\[ \int {\dfrac{1}{3}{{\sin }^2}(u)du} \]
Now we have to solve this integral with respect to “u” and change the “sinx” term in “cosx” because without changing we can not solve the question.
We get;
\[\int {\dfrac{1}{3}{{\sin }^2}(u)du} \\
\Rightarrow \cos 2x = 1 - 2{\sin ^2}x \\
\Rightarrow {\sin ^2}x = \dfrac{{1 - \cos 2x}}{2} \\
\Rightarrow {\sin ^2}u = \dfrac{{1 - \cos 2u}}{2} \\ \]
Using this in our Question, we get
\[\Rightarrow \,\int {\dfrac{1}{3}\left( {\dfrac{{1 - \cos 2u}}{2}} \right)du} \\
\Rightarrow \dfrac{1}{6}\int {(1 - \cos 2u)\,du} \\
\Rightarrow \dfrac{1}{6}\left[ {\int {1du - \left( {\int {\cos 2u\,du} } \right)} } \right] \\
\Rightarrow \dfrac{1}{6}\left[ {u - \dfrac{1}{2}\sin 2u} \right] + C \\
\Rightarrow \dfrac{1}{6}\left[ {3x - \dfrac{1}{2}\sin 2(3x)} \right] + C \\
\Rightarrow \dfrac{1}{6}\left[ {3x - \dfrac{1}{2}\sin 6x} \right] + C \\
\Rightarrow \dfrac{{3x}}{6} - \dfrac{1}{{6 \times 2}}\sin 6x\, + C \\
\therefore \dfrac{x}{2} - \dfrac{1}{{12}}\sin 6x + C \]
This is our required integral.

Note: This is the easiest way to solve this question, you have to, convert the equation in term of “cos x” then solve because there is no any formulae to solve the question directly, you need to break the “sinx” term into single power, without that you can not solve for the power of two on “sinx”. You can check the answer by solving from bottom to top; just you have to differentiate the last equation until you get the first equation and accordingly differentiate the term from bottom.