Question

Find the integral of ${{\cos }^{4}}x$ dx?

Answer 1

As per the given question we have to find the integral of ${{\cos }^{4}}x$ which is quite easy to evaluate using various identities of cos function and then integrating it. Firstly, we will simplify it to the power of one and then integrate it in order to get the answer in simple steps with less chances of errors.

Complete step by step answer:
In the given question we need to find the integral of the cosine function raised to power four which is ${{\cos }^{4}}x$ and also we will make use of the fact that the integration of cosx is sinx and 1dx is x.
Now, in order to integrate the given cosine function what we need to do is use the cosine identity as follows:
${{\cos }^{2}}x=\dfrac{1+\cos 2x}{2}$
So, now we need to integrate $\int{{{\cos }^{4}}xdx}$ and applying the identity we get,
$\begin{aligned} & I=\int{{{\left( \dfrac{1+\cos 2x}{2} \right)}^{2}}dx} \\\ & \Rightarrow \dfrac{1}{4}\int{\left( 1+{{\cos }^{2}}2x+2\cos 2x \right)dx} \\\ \end{aligned}$
Now, applying the identity again and integrating using the fact that integration of cosx is sinx and 1dx is x we get,
$\begin{aligned} & \dfrac{1}{4}\int{\left( 1+{{\cos }^{2}}2x+2\cos 2x \right)dx} \\\ & \Rightarrow \dfrac{1}{4}\left( x+\sin 2x+\int{\left( \dfrac{1+\cos 4x}{2} \right)dx} \right) \\\ & \Rightarrow \dfrac{1}{4}\left( x+\sin 2x+\dfrac{1}{2}x+\dfrac{\sin 4x}{8} \right) \\\ \end{aligned}$
Now, simplifying it further we get,
$\dfrac{1}{4}\left( \sin 2x+\dfrac{3}{2}x+\dfrac{\sin 4x}{8} \right)$
Therefore, the integral of the given cosine function ${{\cos }^{4}}x$dx we get $\dfrac{1}{4}\left( \sin 2x+\dfrac{3}{2}x+\dfrac{\sin 4x}{8} \right)$.

Note: We can also do the given question using a reduction formula and attain the answer but in this we may get the complex step and attain the wrong answer. Also, what we need to do is not to get confused in the integration and differentiation of cos and sine functions.