first uv rule on whole n then again inside for X. cosx... u will get it

Introduction:

To solve the integration of e^x(x.cosx sinx) dx, we will use the technique of integration by parts. This method allows us to integrate a product of two functions by applying the product rule in reverse. The formula for integration by parts is as follows:

∫ u dv = uv - ∫ v du

Where u and v are functions of x, and du and dv are their respective differentials.

Step 1: Identify u and dv:
In our case, we can select u = x and dv = e^x(cosx sinx) dx.

Step 2: Calculate du and v:
To find du, we differentiate u with respect to x, resulting in du = dx.
To find v, we need to integrate dv. This requires simplifying the expression e^x(cosx sinx). We can use the trigonometric identities sin2x = 2sinxcosx and cos2x = cos^2x - sin^2x.

Let's rewrite dv:
dv = e^x(cosx sinx) dx
= e^x(sin2x/2) dx
= (e^x/2)(sin2x) dx
= (e^x/2)(2sinxcosx) dx
= e^x sinx cosx dx

Integrating dv, we obtain:
v = ∫ e^x sinx cosx dx

Step 3: Apply integration by parts:
Using the formula for integration by parts, we have:
∫ u dv = uv - ∫ v du

Applying this to our problem, we get:
∫ x e^x sinx cosx dx = x ∫ e^x sinx cosx dx - ∫ (∫ e^x sinx cosx dx) dx

Simplifying, we have:
I = x ∫ e^x sinx cosx dx - ∫ I dx
I = x ∫ e^x sinx cosx dx - ∫ x ∫ e^x sinx cosx dx dx

Step 4: Solve for the unknown integral:
To solve the integral ∫ e^x sinx cosx dx, we can use integration by parts again.

Let's identify u and dv:
u = sinx, dv = e^x cosx dx

Calculating du and v:
du = cosx dx
v = ∫ e^x cosx dx

We can integrate v using integration by parts:
∫ e^x cosx dx = e^x sinx - ∫ e^x sinx dx

Simplifying, we have:
∫ e^x cosx dx = e^x sinx - ∫ e^x sinx dx

Substituting this back into our main equation, we get:
I = x(e^x sinx - ∫ e^x sinx dx) - ∫ x(e^x sinx cosx dx) dx

Step 5: Simplify and finalize the integral:
We now have an equation with two unknown integrals. To solve for I, we need to simplify and rearrange the terms.

Expanding the equation, we have:
I = x(e^x sinx) -