Question

What is the integral of ${{\cos }^{2}}2x?$

Answer 1

Hint: We will use the trigonometric identities to find the integral. First, we will change the given trigonometric function to an equivalent simple form. And then, we will integrate the function with respect to $x.$ We will use the identity ${{\cos }^{2}}x=\dfrac{1+\cos 2x}{2}.$

Complete step by step solution:
Let us consider the given trigonometric function ${{\cos }^{2}}2x.$
We are asked to find the integral of the given function with respect to $x.$
That is, we need to find the value of $\int{{{\cos }^{2}}x}dx.$
We know the trigonometric identity given by ${{\cos }^{2}}x=\dfrac{1+\cos 2x}{2}.$
In this identity, we will put $2x$ instead of $x,$ since our function contains $2x.$
We will get ${{\cos }^{2}}\left( 2x \right)=\dfrac{1+\cos 2\left( 2x \right)}{2}.$
We can write this as \[{{\cos }^{2}}2x=\dfrac{1+\cos 4x}{2}.\]
Now, we can change the integral using the above equation.
We will get $\int{{{\cos }^{2}}2x}dx=\int{\dfrac{1+\cos 4x}{2}dx}.$
Let us take $\dfrac{1}{2}$ out of the integral sign, since it is a constant term. We know that we can take the constant term if it is multiplied with a variable term. If the operation is addition, then we cannot take the term out.
Now, we will get $\int{{{\cos }^{2}}2x=\dfrac{1}{2}\int{\left( 1+\cos 4x \right)dx}.}$
Now, we can use the linearity property of integration.
We will get $\int{{{\cos }^{2}}2x}=\dfrac{1}{2}\left\{ \int{1dx}+\int{\cos 4xdx} \right\}.$
Let us consider the first integral, $\int{1}dx=x.$
We know that the integral of the Cosine function is the Sine function.
So, from the second integral, we will get $\int{\cos 4xdx}=\dfrac{\sin 4x}{4}.$
So, the integral will become $\int{{{\cos }^{2}}2x}dx=\dfrac{1}{2}\left\{ x+\dfrac{\sin 4x}{4} \right\}+C.$
Hence the integral of the given function is $\int{{{\cos }^{2}}2x}dx=\dfrac{x}{2}+\dfrac{\sin 4x}{8}+C.$

Note: Remember the similar trigonometric identity ${{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}.$ By applying this identity, we can find the integral of ${{\sin }^{2}}2x$ in the exact way we have found the integral of ${{\cos }^{2}}2x.$ The linearity property of integration is $\int{\left( af\left( x \right)+bg\left( x \right) \right)dx=a\int{f\left( x \right)dx}+b\int{g\left( x \right)dx}}$ where $a,b$ are constants and $f,g$ are functions of $x.$