English

Evaluate : 2 π ∫ 0 Cos 5 X D X .

Advertisements
Advertisements

Question

Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .

Advertisements

Solution

Let I =`int_0^(2x) cos ^5 x  dx` ..... (1)

` cos^5( 2π - x )= cos^5 x `

It is known that,

`int_0^(2a) f (x) dx = 2 int _0^a f (x)dx, if f (2a - x ) = f(x)`

                      = 0 if f (2a - x = - f (x)

∴ `I = 2 int_0^π cos^5 x  dx `

Now 

\[f\left( \pi - x \right) = \cos^5 \left( \pi - x \right) = - \cos^5 x = - f\left( x \right)\]

⇒ I = 2(0 ) = 0                 [ `cos^5(π - x) = - cos^5 x`]                             

shaalaa.com
  Is there an error in this question or solution?
2016-2017 (March) Foreign Set 3

RELATED QUESTIONS

\[\int\limits_{\pi/6}^{\pi/4} cosec\ x\ dx\]

\[\int\limits_{\pi/3}^{\pi/4} \left( \tan x + \cot x \right)^2 dx\]

\[\int\limits_0^{\pi/2} x^2 \cos\ 2x\ dx\]

\[\int\limits_0^2 \frac{1}{4 + x - x^2} dx\]

\[\int\limits_0^1 \frac{1}{2 x^2 + x + 1} dx\]

\[\int_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\]

\[\int_0^\frac{\pi}{4} \left( a^2 \cos^2 x + b^2 \sin^2 x \right)dx\]

\[\int\limits_1^2 \frac{3x}{9 x^2 - 1} dx\]

\[\int\limits_0^1 \frac{\tan^{- 1} x}{1 + x^2} dx\]

\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{3/2}} dx\]

\[\int\limits_{- a}^a \sqrt{\frac{a - x}{a + x}} dx\]

\[\int_0^\frac{1}{2} \frac{1}{\left( 1 + x^2 \right)\sqrt{1 - x^2}}dx\]

\[\int_0^\frac{\pi}{2} \sqrt{\cos x - \cos^3 x}\left( \sec^2 x - 1 \right) \cos^2 xdx\]

\[\int_0^\frac{\pi}{2} \frac{\cos x}{\left( \cos\frac{x}{2} + \sin\frac{x}{2} \right)^n}dx\]

\[\int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx\]

 


\[\int\limits_2^3 \left( 2 x^2 + 1 \right) dx\]

\[\int\limits_1^2 \left( x^2 - 1 \right) dx\]

\[\int\limits_1^4 \left( 3 x^2 + 2x \right) dx\]

\[\int\limits_0^2 \left( x^2 + 2x + 1 \right) dx\]

\[\int\limits_a^b x\ dx\]

\[\int\limits_{- \pi/2}^{\pi/2} \sin^3 x\ dx .\]

\[\int\limits_0^2 \sqrt{4 - x^2} dx\]

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{2} e^x \left( \sin x - \cos x \right)dx\]

 


The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .


\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]


\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]


\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]


Evaluate the following using properties of definite integral:

`int_0^1 log (1/x - 1)  "d"x`


Evaluate the following:

`int_0^oo "e"^(-mx) x^6 "d"x`


Evaluate the following:

`int_0^oo "e"^(-4x) x^4  "d"x`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×