Question

How to find integral of \[{\cos ^n}x\]?

Answer 1

Hint: Finding integral is just the opposite of differentiation so for finding an integral you have to go reverse as, in differentiation, it's quite easy when you get it. In this question, you have to buy the parts method and you will have to use a few trigonometric identities as per requirement in the solution.

Complete step by step answer:
calculating the integral with \[dx\]we get:
$\Rightarrow$ \[\int {{{\cos }^n}x} dx = \sin x{\cos ^{n - 1}}x - \int {(\sin x)(n - 1){{\cos }^{n - 1}}x( - \sin x)} \,\]
Now, we know \[{\sin ^2}x = 1 - {\cos ^2}x\] to get \[\,{\sin ^2}x\]in cosine term
$\Rightarrow$ \[\int {{{\cos }^n}x} dx = \]\[\sin x{\cos ^{n - 1}}x + (n - 2)\{ \sin x{\cos ^{n - 2}}x - \int {(\sin x)(n - 2){{\cos }^{n - 2}}x( - \sin x)\} } - (n - 1)\{ \sin x{\cos ^{n - 1}}x - \int {(\sin x)(n - 1){{\cos }^{n - 1}}x( - \sin x)\} } \]
Let \[\sin x{\cos ^{n - 1}}x - \int {(\sin x)(n - 1){{\cos }^{n - 1}}x( - \sin x)} \] \[ = {I_n}\] now replacing \[{I_n}\]in above equation we get,
$\Rightarrow$ \[\int {{{\cos }^n}x} dx = \]\[\sin x{\cos ^{n - 1}}x + (n - 1)\{ {I_{n - 2}}\} - (n - 1){I_n}\]
$\Rightarrow$ \[\int {{{\cos }^n}x} dx + (n - 1){I_n} = \sin x{\cos ^{n - 1}}x + (n - 1)\int {{{\cos }^{n - 2}}x} dx\]
$\Rightarrow$ \[\int {{{\cos }^n}x} dx + n\int {{{\cos }^n}x} dx - 1(\int {{{\cos }^n}x} dx) = \sin x{\cos ^{n - 1}}x + (n - 1)\int {{{\cos }^{n - 2}}x} dx\]
$\Rightarrow$ \[n\int {{{\cos }^n}x} dx = \sin x{\cos ^{n - 1}}x + (n - 1)\int {{{\cos }^{n - 2}}x} dx\]
Dividing \[n\] both sides the final answer is obtained
$\Rightarrow$ \[\int {{{\cos }^n}x} dx = \dfrac{1}{n}\sin x{\cos ^{n - 1}}x + \dfrac{1}{n}(n - 1)\int {{{\cos }^{n - 2}}x} dx\]
Formulae Used: By parts, and simple trigonometric identity are used.
Additional Information: You can check the answer by solving from bottom to top; just you have to differentiate the last equation until you get the first equation. This method can take some time and several steps more than you take for integration but definitely, you can get and verify.

Note:
This is the easiest way to solve this question, but if you find any difficulty then you can try another method that is, use an expansion of \[{\cos ^n}x\]and after expansion you can integrate each term independently, then you can merge it and obtain your result.
Expansion method is not so reliable because you can’t learn expansion of each term what you asked for and you even don’t know which question can appear in front of you, so use the best approach that is explained above and get your results with such question in a short and accurate way.