Question

What is the integral of $\int{{{\sin }^{4}}\left( 4x \right)dx}$?

Answer 1

First of all write the function inside the integral as ${{\sin }^{4}}\left( 4x \right)={{\left( {{\sin }^{2}}\left( 4x \right) \right)}^{2}}$. Use the trigonometric identity ${{\sin }^{2}}\theta =\dfrac{1-\cos 2\theta }{2}$ and simplify this function. Now, break the integral into parts by using the algebraic identity ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$. Apply the formula ${{\cos }^{2}}\theta =\dfrac{1+\cos 2\theta }{2}$ for further simplification. Finally, use the formulas $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ and $\int{\cos \left( ax+b \right)dx}=\dfrac{\sin \left( ax+b \right)}{a}$ to evaluate the integral. Add the constant of indefinite integral (c) in the end.

Complete step by step solution:
Here we are asked to find the integral of the function ${{\sin }^{4}}\left( 4x \right)$. First let us simplify the trigonometric function using the half angle formula. Let us assume the integral as I, so we have,
$\Rightarrow I=\int{{{\sin }^{4}}\left( 4x \right)dx}$
We can write ${{\sin }^{4}}\left( 4x \right)={{\left( {{\sin }^{2}}\left( 4x \right) \right)}^{2}}$, so we get,
$\Rightarrow I=\int{{{\left( {{\sin }^{2}}\left( 4x \right) \right)}^{2}}dx}$
Using the half angle trigonometric identity given as ${{\sin }^{2}}\theta =\dfrac{1-\cos 2\theta }{2}$ we get,

& \Rightarrow I=\int{{{\left( \dfrac{1-\cos 8x}{2} \right)}^{2}}dx} \\\ & \Rightarrow I=\dfrac{1}{4}\int{{{\left( 1-\cos 8x \right)}^{2}}dx} \\\ \end{aligned}$$ Using the algebraic identity ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ and breaking the integral into parts we get, $$\begin{aligned} & \Rightarrow I=\dfrac{1}{4}\int{\left( 1+{{\cos }^{2}}\left( 8x \right)-2\cos \left( 8x \right) \right)dx} \\\ & \Rightarrow I=\dfrac{1}{4}\left[ \int{1dx}+\int{{{\cos }^{2}}\left( 8x \right)dx}-\int{2\cos \left( 8x \right)dx} \right] \\\ \end{aligned}$$ Here, 1 can be written as ${{x}^{0}}$ so applying the formulas $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$, $$\int{\cos \left( ax+b \right)dx}=\dfrac{\sin \left( ax+b \right)}{a}$$ we get, $$\begin{aligned} & \Rightarrow I=\dfrac{1}{4}\left[ \dfrac{{{x}^{0+1}}}{0+1}+\int{{{\cos }^{2}}\left( 8x \right)dx}-2\times \dfrac{\sin \left( 8x \right)}{8} \right] \\\ & \Rightarrow I=\dfrac{1}{4}\left[ x+\int{{{\cos }^{2}}\left( 8x \right)dx}-\dfrac{\sin \left( 8x \right)}{4} \right] \\\ \end{aligned}$$ Now, using the half angle trigonometric identity ${{\cos }^{2}}\theta =\dfrac{1+\cos 2\theta }{2}$ we get, $$\begin{aligned} & \Rightarrow I=\dfrac{1}{4}\left[ x+\int{\dfrac{1+\cos \left( 16x \right)}{2}dx}-\dfrac{\sin \left( 8x \right)}{4} \right] \\\ & \Rightarrow I=\dfrac{1}{4}\left[ x+\dfrac{1}{2}\int{1dx}+\dfrac{1}{2}\int{\cos \left( 16x \right)dx}-\dfrac{\sin \left( 8x \right)}{4} \right] \\\ \end{aligned}$$ Again using the formulas $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$, $$\int{\cos \left( ax+b \right)dx}=\dfrac{\sin \left( ax+b \right)}{a}$$ we get, $$\begin{aligned} & \Rightarrow I=\dfrac{1}{4}\left[ x+\dfrac{x}{2}+\dfrac{1}{2}\times \dfrac{\sin \left( 16x \right)}{16}-\dfrac{\sin \left( 8x \right)}{4} \right] \\\ & \Rightarrow I=\dfrac{1}{4}\left[ \dfrac{3x}{2}+\dfrac{\sin \left( 16x \right)}{32}-\dfrac{\sin \left( 8x \right)}{4} \right] \\\ & \Rightarrow I=\dfrac{1}{128}\left[ 48x+\sin \left( 16x \right)-8\sin \left( 8x \right) \right]+c \\\ \end{aligned}$$ Here ‘c’ is the constant of integration as we are evaluating an indefinite integral. Hence, the above relation is our answer. **Note:** Note that we don’t have any direct formula for the integration of the function ${{\sin }^{4}}\left( 4x \right)$or even for the function ${{\sin }^{2}}\left( 4x \right)$ so it must be converted into the cosine function using the half angle formula. Also, remember that you cannot apply the formula $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ for n = -1 because in that case we have the function $\dfrac{1}{x}$ whose integral is $\ln x$. Remember the formulas of integral and differential of all the trigonometric functions.