How do you find integral of cosn x?
Sub Grade: 10

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answer is C.

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Detailed Solution

cosn x =

(\frac{1}{n}) \cos^{n - 1} x sin x + \frac{n - 1}{n} \cos^{n - 2} xdx

Let I n=

sin x  cos 6^{n - 1} x - (sinx) (n - 1) co s^{n - 1} x (- sinx) dx

Followed by using

\sin^{2} x = 1 - co s^{2} x

to get the sin2 back to a cosine.
But this gives you (n-1) cosn xdx some where on the right.

In = sin x co s^{n - 1} x + (n - 1) I n - 2 - (n - 2) In

n cosn xdx =

sin x co s^{n - 1} x + (n - 1) co s^{n - 2} xdx

Dividing through by n gives the reduction formula
For any particular small positive value of n you can apply this repeatedly to down to the integral either of 1 or cos x.
Type Exam: CBSE

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