In=∫ x^nSinx dx....with the limits π/2 and 0
i found that In would be equal to n(π/2)^(n-1) -n(n-1) In-2
how do i deduce the value of I4? where n=4
you need to do integration by parts.
∫x^4 sinx dx
u=x^4
du = 4x^3 dx
dv = sinx dx
v = -cosx
∫x^4 sinx = -x^4 cosx + 4∫x^3 cosx dx
Now do all that again for each resulting integral, and the powers of x descends until you are left with just ∫sinx dx
Then just add all the pieces together.
Hmmm. I think you can save yourself some work. At x=0 all the terms with powers of x and sinx vanish.
at x=π/2 all the terms with cosx vanish (odd powers of x).
To find the value of I4, which corresponds to the integral ∫ x^4sinx dx with limits π/2 and 0, you can use the recurrence relation formula you mentioned:
In = n(π/2)^(n-1) - n(n-1)In-2
where In represents the integral ∫ x^nSinx dx.
To deduce the value of I4, let's substitute n=4 into the formula and solve step by step:
I4 = 4(π/2)^(4-1) - 4(4-1)I2
First, simplify the exponent:
I4 = 4(π/2)^3 - 4(3)I2
Next, evaluate (π/2)^3:
I4 = 4(π/8) - 4(3)I2
Simplify further:
I4 = π/2 - 12I2
Now, we need to find the value of I2. To do this, we can use the same recurrence relation formula:
I2 = 2(π/2)^(2-1) - 2(2-1)I0
Simplify:
I2 = 2(π/2) - 2I0
The integral I0 corresponds to ∫ x^0sinx dx, which is simply the integral of sinx over the given limits.
To evaluate I0, integrate sinx with respect to x:
∫ sinx dx = -cosx
Evaluate the antiderivative at the limits:
I0 = -cos(π/2) - (-cos(0))
Simplify:
I0 = -0 + 1
I0 = 1
Now, substitute I0 = 1 into the formula for I2:
I2 = 2(π/2) - 2(2-1)(1)
Simplify:
I2 = π - 2
Finally, substitute I2 = π - 2 into the formula for I4:
I4 = π/2 - 12(π - 2)
Simplify:
I4 = π/2 - 12π + 24
Therefore, the value of I4 is π/2 - 12π + 24.