Homework Help

Explain how is calculate definite integral (upp lim pi-low lim 0) sin^5xdx?

user profile pic

sparangello | Student, College Freshman | eNoter

Posted August 26, 2012 at 12:54 PM via web

dislike 3 like

Explain how is calculate definite integral (upp lim pi-low lim 0) sin^5xdx?

Tagged with calculate, definite, integral, math, pi

2 Answers | Add Yours

Top Answer

user profile pic

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted August 26, 2012 at 1:37 PM (Answer #1)

dislike 2 like

`int^pi_0 sin^5x dx`

Let;

t = cosx

dt = -sinx dx

 


When x = 0 then t = cos0 = 1

When x = `pi ` then t = `cospi` = -1

 

`int^pi_0 sin^5x dx`

`= int^pi_0 (sin^2x)^2.sinx dx`

`= int^(-1)_1 (1-t^2)^2-dt`

 

The limits of integral go across 0. So we need to integrate this as addition of two parts.

`int^(-1)_1 (1-t^2)^2-dt`

`=-[int^(0)_1(1-t^2)^2dt+int^(-1)_0 (1-t^2)^2 dt]`

`= -[int^(-1)_0 (1-2t^2+t^4) dt + int^0_1 (1-2t^2+t^4) dt]`

`= -{[t-2t^3/3+t^5/5]^(-1)_0 + [t-2t^3/3+t^5/5]^0_1] }`

`= 16/15`

 

`int^pi_0 sin^5x dx` = 16/15

Sources:

user profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted August 26, 2012 at 1:37 PM (Answer #2)

dislike 2 like

I suggest to you to deduce a reccurence relation to help you to evaluate the integral easier.

You should  come up with the following notation such that:

`I_n = int_0^pi sin^n x dx and I_1 = int_0^pi sin x dx`

You may evaluate `I_1`  such that:

`I_1 = -cos x|_0^pi => I_1 = -cos pi + cos 0 => I_1 = -(-1)+1 = 2`

You may write `I_n`  such that:

`I_n = int_0^pi sin^(n-1) x *sin xdx`

Notice that `sin x dx = (-cos x)' dx`  such that:

`I_n = int_0^pi sin^(n-1) x *(-cos x)' dx`

Using integration by parts yields:

`f(x) = sin^(n-1) x => f'(x) = (n-1)sin^(n-2)x*cos x`

`g'(x) = sin x => g(x) = -cos x`

`I_n =-cos x*sin^(n-1) x|_0^pi + (n-1)int_0^pi sin^(n-2)x*cos^2 x dx`

Substituting `1-sin^2 x`  for `cos^2 x`  yields:

`I_n =-cos x*sin^(n-1) x|_0^pi + (n-1)int_0^pi sin^(n-2)x*(1-sin^2 x)dx`

`I_n =-cos x*sin^(n-1) x|_0^pi + (n-1)int_0^pi sin^(n-2)x dx - (n-1)int_0^pi sin^n x dx`

But `int_0^pi sin^n x dx = I_n,`  hence, substituting `I_n`  for int_0^pi sin^n x dx yields:

`I_n = -cos x*sin^(n-1) x|_0^pi + (n-1)int_0^pi sin^(n-2)x dx - (n-1)I_n`

`nI_n= 0 - 0 + (n-1)I_(n-2)`

`I_n = ((n-1)/n)I_(n-2)`

Since you know the equation that relates `I_n`  and `I_(n-2), ` you may evaluate `I_5`  such that:

`I_5 = (5-1)/5*I_3 => I_5 = (5-1)/5*(3-1)/3*I_1`

Since `I_1 = 2` , you may evaluate `I_5`  such that:

`I_5 = (4/5)(2/3)*2 => I_5 = 16/15`

Hence, evaluating the definite integral using the reccurence relation `I_n = ((n-1)/n)I_(n-2)`  yields `I_5 = 16/15` .

Sources:

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes