Question

integration of 0 to pi cos cube x dx​

Answer 1

Hey mate plzz refer to the attachment

Plzz check ur question again mistake is pi /2(limit) ,

Answer 2

Answer:

The value of given integral is

$I=\int _0^{\pi }cos^3xdx$

Step-by-step explanation:

The correct question is"Find the integral of cos^{3}xdx under the limits from 0 to π"

Let the given integral be

$I=\int _0^{\pi }cos^3xdx$

The above integral can be rearranged as given below

$I=\int _0^{\pi }cosx.cos {}^{2}x dx \\ = \int _0^{\pi }cosx(1 - cos {}^{2} x)dx \\ = \int _0^{\pi }cosx.sin {}^{2} x \: dx$

We solve this integral by substitution method as shown below

$let \: sinx = t \\ = > cosx \: dx = dt$

x ranges fro 0 to π then t ranges from 0 to 0.Hence,the integral becomes

$I = \int _0^{0 }t {}^{2} dt \\ = (\frac{t {}^{3} }{3} )_0^{0 } = 0 - 0 = 0$

Therefore,the value of given integral is

$I=\int _0^{\pi }cos^3xdx = 0$

#SPJ3