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

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`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:**

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:**

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