Question

find integral of cos³xdx​

Answer 1

Answer:

$\sin(x) - \frac{ \sin ^{3} x}{3} + c$

where c is integral constant .

Step-by-step explanation:

change cos^3x =cos^2x•cosx

again change cos^2x=1-sin^2x

and now put the value in the given ques .

then use substitution .

let sinx=t

= cosxdx=dt

and put the value ...

solution refer to the attachment

Answer 2

The integra value of cos³xdx​ is  $\frac{1}{4}$​[$\frac{sin3x}{3}$​+3sinx]+C.

Given:

cos³xdx​

To mind:

We have to find the integral of cos³xdx​.

Solution:

Consider

I=∫cos3x dx

II=∫$\frac{(cos3x+3cosx dx)}{4}$)      [ We know cos3A=4cos3A−3cosA ]

I=$\frac{1}{4}$∫cos3x+3cosx dx

I=$\frac{1}{4}$​∫cos3x dx+$\frac{1}{4}$∫3cosx dx

I=$\frac{1}{4}$​[$\frac{sin3x}{3}$​+3sinx]+C

(Where C is a constant.)

So the integra value of cos³xdx​ is  $\frac{1}{4}$​[$\frac{sin3x}{3}$​+3sinx]+C.

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