Question

Solve the following integral:
 $\int {{{\cos }^2}nxdx} $

Answer 1

Hint: Before attempting this question one must have prior knowledge of the method of integration and formula used in this method such as\[\int {{x^n}} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C\] use this information to approach the solution of the problem.

Complete step by step solution:
So by the given information we have function $\int {{{\cos }^2}nxdx} $
Let’s use the method of integration
I =$\int {{{\cos }^2}nxdx} $ (equation 1)
Let nx = t
Differentiating both side with respect to x
$n\dfrac{d}{{dx}}x = \dfrac{{dt}}{{dx}}$
Since we know that $\dfrac{dx}{{dx}}$ = 1
Therefore n = $\dfrac{{dt}}{{dx}}$
$ \Rightarrow $$dx = \dfrac{{dt}}{n}$
Substituting the value of dx in equation 1 we get
I =$\int {{{\cos }^2}t\dfrac{{dt}}{n}} $
Since we know that according to the trigonometric identity of $\cos \theta $ i.e. $\cos 2\theta = 2{\cos ^2}\theta - 1$
Simplifying $\cos 2\theta = 2{\cos ^2}\theta - 1$ we get
 \[{\cos ^2}\theta = \dfrac{{\cos 2\theta + 1}}{2}\]
By the above equation I =\[\dfrac{1}{n}\int {\dfrac{{1 + \cos 2t}}{2}dt} \]
$ \Rightarrow $ I =\[\dfrac{1}{{2n}}\left( {\int {1dt} + \int {\cos 2tdt} } \right)\] (equation 2)
Let 2t = q
Differentiating both side with respect to t
$2\dfrac{{dt}}{{dt}} = \dfrac{{dq}}{{dt}}$
Since we know that $\dfrac{dt}{{dt}}$ = 1
$ \Rightarrow $$2 = \dfrac{{dq}}{{dt}}$
$ \Rightarrow $$dt = \dfrac{{dq}}{2}$
Substituting the value of dt in equation 2
I =\[\dfrac{1}{{2n}}\left( {\int {1dt} + \int {\cos q\dfrac{{dq}}{2}} } \right)\]
I =\[\dfrac{1}{{2n}}\int {1dt} + \dfrac{1}{{2n}}\int {\cos q\dfrac{{dq}}{2}} \] (equation 3)
For \[\dfrac{1}{{2n}}\int {1dt} \] integrating with respect to t
$ \Rightarrow $\[\dfrac{1}{{2n}}t\]
Now for \[\dfrac{1}{{2n}}\int {\cos q\dfrac{{dq}}{2}} \] integrating with respect to q we get
$ \Rightarrow $\[\dfrac{1}{{4n}}\sin q\]
Substituting the values in the equation 3 we get
I =\[\dfrac{1}{{2n}}t + \dfrac{1}{{4n}}\sin q\] + C
Now substituting the value of q in the above equation
I =\[\dfrac{1}{{2n}}t + \dfrac{1}{{4n}}\sin 2t\] + C
Now substituting the value of t in the above equation we get
I =\[\dfrac{1}{{2n}}nx + \dfrac{1}{{4n}}\sin 2nx\] + C
$ \Rightarrow $ I =\[\dfrac{x}{2} + \dfrac{{\sin 2nx}}{{4n}}\] + C
Hence $\int {{{\cos }^2}nxdx} $ = \[\dfrac{x}{2} + \dfrac{{\sin 2nx}}{{4n}}\] + C.

Note: In the above solution we used the method of integration which can be explained as the method which is used to add the small slices of whole body it is also used to find the volume, area of irregular shapes body also it is collection of small pieces of data there is a term used in the method of integration i.e. limit which is used to help that how close we are to collect all the small pieces of data.