Question

What is the integral of ${{\left( \cos x \right)}^{2}}$?

Answer 1

From the given question we have to find the integral of ${{\left( \cos x \right)}^{2}}$. To solve the above question we will use the cosine double angle identity in order to rewrite ${{\left( \cos x \right)}^{2}}$ as ${{\cos }^{2}}x$, as we know that the $\cos 2x=2{{\cos }^{2}}x-1$ by this we will get ${{\cos }^{2}}x=\dfrac{\cos 2x+1}{2}$ in place of ${{\cos }^{2}}x$ we will write this and we will integrate.

Complete step by step solution:
From the given question we have to find the integral of
${{\left( \cos x \right)}^{2}}$
To solve the above question, we will use the cosine double angle identity in order to rewrite ${{\left( \cos x \right)}^{2}}$ as
$\Rightarrow {{\left( \cos x \right)}^{2}}={{\cos }^{2}}x$
As we know that the,
$\Rightarrow \cos 2x=2{{\cos }^{2}}x-1$
by this we will get
$\Rightarrow {{\cos }^{2}}x=\dfrac{\cos 2x+1}{2}$
In place of ${{\cos }^{2}}x$ we will write this and we will integrate now, that means,
Thus,
$\Rightarrow \int{{{\cos }^{2}}x}=\int{\dfrac{\cos 2x+1}{2}}$
Now we will split up the integral,
$\Rightarrow \int{{{\cos }^{2}}x}dx=\int{\dfrac{\cos 2x}{2}}dx+\dfrac{1}{2}\int{dx}$
As we know that second integral is perfect integral, that means,
$\Rightarrow \int{dx}=x+c$
By this we will get,
$\Rightarrow \int{{{\cos }^{2}}x}dx=\int{\dfrac{\cos 2x}{2}}dx+\dfrac{1}{2}x$
The constant of integration will be added upon evaluating the remaining integral.
Now for the Cosine integral, we will use substitution,
Let
$\Rightarrow u=2x$
after differentiating on both sides, we will get,
$\Rightarrow du=2dx$
$\Rightarrow \dfrac{1}{2}du=dx$
Now substitute the above in the equation, we will get,
$\Rightarrow \int{{{\cos }^{2}}x}dx=\int{\dfrac{\cos 2x}{2}}dx+\dfrac{1}{2}x$
After substituting, we will get,
$\Rightarrow \int{{{\cos }^{2}}x}dx=\dfrac{1}{4}\int{\cos \left( u \right)}du+\dfrac{1}{2}x$
As we know that integral of Cos is sin, that means,
$\Rightarrow \int{\operatorname{Cos}\left( u \right)=\operatorname{Sin}\left( u \right)+c}$
By this we will get,
$\Rightarrow \int{{{\cos }^{2}}x}dx=\dfrac{1}{4}\sin \left( u \right)+\dfrac{1}{2}x+c$
Since $\Rightarrow u=2x$
By this we will get,
$\Rightarrow \int{{{\cos }^{2}}x}dx=\dfrac{1}{4}\sin \left( 2x \right)+\dfrac{1}{2}x+c$
Therefore, this is the required answer.

Note: Students should recall all the formulas of trigonometry and integration, formulas like
$\begin{aligned} & \Rightarrow \int{\cos \left( u \right)du=\sin \left( u \right)+c} \\\ & \Rightarrow \cos 2x=2{{\cos }^{2}}x-1=1-2{{\sin }^{2}}x={{\cos }^{2}}x-{{\sin }^{2}}x \\\ & \Rightarrow \operatorname{Sin}2x=2\operatorname{Sin}x\operatorname{Cos}x \\\ & \Rightarrow \int{\sin xdx}=-\cos x+c \\\ & \Rightarrow \int{{{\sec }^{2}}xdx}=\tan x+c \\\ & \Rightarrow \int{dx}=x+c \\\ \end{aligned}$
Students should not forget to write the plus constant “C” at the end of the solution.