First make an appropriate substitution and then use integration by parts to evaluate the indefinite integrals:
∫ sinx cos³x e^(1-sin²x) dx
I was going to substitute u= 1 - sin²x, but then i got du = ½ sinxcosx - ½ xdx, so im stuck.
Thank you so much!!
Isn't (1-sin²x)=cos²x?
You will end up with an expression containing cos(x) and ONE sin(x).
Try u=cos(x), du=d(cos(x))=-sin(x)dx
Do you see the light?
To evaluate the indefinite integral ∫ sinx cos³x e^(1-sin²x) dx, you started off with the substitution u = 1 - sin²x. This was a good choice because it simplifies the expression inside the exponential function.
Now, let's proceed with the integration by parts. The formula for integration by parts is:
∫ u dv = uv - ∫ v du
In this case, let's choose the following substitutions:
u = cos³x (choose this as the 1st function because its derivative is simpler)
dv = e^(1-sin²x) dx (choose this as the 2nd function)
Next, we need to find du and v. Taking the derivative of u = cos³x:
du/dx = -3cos²x sinx
Integrating dv = e^(1-sin²x) dx requires a substitution. Substitute t = sinx:
v = ∫ e^(1-sin²x) dx
Let t = sinx, then dt = cosx dx
v = ∫ e^(1-t²) dt
Now, we have all the necessary components to apply the integration by parts formula:
∫ sinx cos³x e^(1-sin²x) dx = ∫ u dv
= uv - ∫ v du
= cos³x ∫ e^(1-sin²x) dx - ∫ (∫ e^(1-t²) dt) (-3cos²x sinx)
Now we have two integrals to evaluate: ∫ e^(1-sin²x) dx and ∫ (∫ e^(1-t²) dt) (-3cos²x sinx). Let's start with the first one.
∫ e^(1-sin²x) dx:
Since u = 1 - sin²x, substitute back to the original variable:
u = 1 - sin²x
du = -2 sinx cosx dx
The integral becomes:
∫ e^(1-sin²x) dx = ∫ e^u du/(-2 sinx cosx)
= (-1/2) ∫ e^u du
∫ e^(1-sin²x) dx = (-1/2) e^u + C1
= (-1/2) e^(1-sin²x) + C1
Now, move on to the second integral:
∫ (∫ e^(1-t²) dt) (-3cos²x sinx)
To evaluate this integral, we need to find the integral of e^(1-t²) with respect to t:
∫ e^(1-t²) dt:
This integral does not have a simple closed-form expression, so it cannot be evaluated analytically. It requires numerical methods or special functions to evaluate it.
However, since we are interested in evaluating ∫ (∫ e^(1-t²) dt) (-3cos²x sinx), we can simplify this expression by substituting back t = sinx:
∫ (∫ e^(1-t²) dt) (-3cos²x sinx)
= ∫ (∫ e^(1-sin²x) dt) (-3cos²x sinx)
= ∫ (∫ e^(1-sin²x) dx) (-3cos²x sinx)
We already found that ∫ e^(1-sin²x) dx = (-1/2) e^(1-sin²x) + C1, so the expression becomes:
= ∫ (-1/2) e^(1-sin²x) dx (-3cos²x sinx)
= (-3/2) (∫ e^(1-sin²x) dx) (cos²x sinx)
= (-3/2) ((-1/2) e^(1-sin²x) + C1) (cos²x sinx)
Finally, combining the results obtained:
∫ sinx cos³x e^(1-sin²x) dx = cos³x ∫ e^(1-sin²x) dx - ∫ (∫ e^(1-t²) dt) (-3cos²x sinx)
= cos³x (-1/2) e^(1-sin²x) + C1 - (-3/2) ((-1/2) e^(1-sin²x) + C1) (cos²x sinx)
= (-1/2) cos³x e^(1-sin²x) + 3/4 e^(1-sin²x) cos²x sinx + C2
Therefore, the indefinite integral is:
∫ sinx cos³x e^(1-sin²x) dx = (-1/2) cos³x e^(1-sin²x) + 3/4 e^(1-sin²x) cos²x sinx + C2
where C1 and C2 are arbitrary constants.