Question

How do you find the integral of $ \int {{{\cos }^2}\theta } $ ?

Answer 1

Hint: In order to find the integral of $ \int {{{\cos }^2}\theta } $ , we will change it into another form using the trigonometric identity i.e., $ \cos 2\theta = 2{\cos ^2}\theta - 1 $ , since this integration cannot be evaluated by the direct formula of integration. Then by evaluating it using integration identities like $ \int {\cos n\theta .d\theta = \dfrac{1}{n}\left( {\sin n\theta } \right)} $ and $ \int 1 .d\theta = \theta $ , we will determine the required integration.

Complete step-by-step answer:
Now, we need to find the integral of $ \int {{{\cos }^2}\theta } $ .
This integration cannot be evaluated by the direct formula of integration. Thus, let us change it into another form using the trigonometric identity.
We know that $ \cos 2\theta = 2{\cos ^2}\theta - 1 $ .
Therefore, $ {\cos ^2}\theta = \dfrac{1}{2}\left( {\cos 2\theta + 1} \right) $
Hence, $ \int {{{\cos }^2}\theta } .d\theta = \int {\dfrac{1}{2}\left( {\cos 2\theta + 1} \right)} .d\theta $
 $ \int {{{\cos }^2}\theta } .d\theta = \dfrac{1}{2}\int {\cos 2\theta .d\theta + \int 1 .d\theta } $
We know that $ \int {\cos n\theta .d\theta = \dfrac{1}{n}\left( {\sin n\theta } \right)} $ and $ \int 1 .d\theta = \theta $
Therefore, $ \int {{{\cos }^2}\theta } .d\theta = \dfrac{1}{2}[\left( {\dfrac{1}{2}\sin 2\theta + \theta } \right]$
 $ \int {{{\cos }^2}\theta } .d\theta = \dfrac{1}{2}\theta + \dfrac{1}{4}\sin 2\theta $
Hence, the integral of $ \int {{{\cos }^2}\theta } $ is $ \dfrac{1}{2}\theta + \dfrac{1}{4}\sin 2\theta $
So, the correct answer is “ $ \dfrac{1}{2}\theta + \dfrac{1}{4}\sin 2\theta $ ”.

Note: The integration denotes summation of discrete data. The integral is calculated to find the functions which will describe the area, displacement, volume, that occur due to a small data, which cannot be measured singularly. The concept of integration has developed to find the problem function, when its derivatives are given. And to find the area bounded by the graph of a function under certain constraints.
Generally, integration is a method of adding or summing up the parts to find the whole. It is a reverse process of differentiation, where we reduce the functions into parts. This method is used to find the summation under a vast scale. Calculation of small addition problems is an easy task which we can do manually or by using calculators as well. But for big addition problems, where limits could reach infinity, integration methods are used. Integration and differentiation both are important parts of calculus.