I ran across this one the other day. See what you can make of it.
Find the integral of
ecosxsinx - (sinx+cosx)esinx+cosx
-----------------------------------------
esin2x-2esinx+1
How about you ask someone to explain it and not give you the answer
To find the integral of the given expression, you need to use integration techniques such as substitution and manipulation of exponential and trigonometric functions. Let's go through the steps of solving this integral.
First, let's simplify the expression:
∫ (e^(cosx)sinx - (sinx + cosx)e^(sinx+cosx)) / (e^(sin2x) - 2e^(sinx) + 1) dx
To get started, notice that the denominator looks like a quadratic binomial. You can rewrite it as the square of a binomial:
(e^(sinx) - 1)^2
Therefore, the integral can be rewritten as:
∫ (e^(cosx)sinx - (sinx + cosx)e^(sinx+cosx)) / (e^(sinx) - 1)^2 dx
Now, let's substitute a new variable to simplify the expression further. Let u = e^(sinx).
Differentiating both sides with respect to x gives: du/dx = e^(sinx)cosx.
Substituting u and du into the integral, we get:
∫ (sinx - (sinx + cosx)(e^(sinx + cosx))) / u^2 * du/dx dx
Simplifying, we have:
∫ (sinx - (sinx + cosx)(e^(sinx + cosx))) / u^2 * e^(sinx)cosx dx
= ∫ (sinx - (sinx + cosx)(e^(sinx + cosx))) / u^2 * e^(sinx)cosx * 1/du/dx du
= ∫ (sinx - (sinx + cosx)(e^(sinx + cosx))) / u^2 * e^(sinx)cosx * du/dx dx/du
At this point, we can simplify the expression further. Notice that du/dx = e^(sinx)cosx, so dx/du = 1 / (e^(sinx)cosx).
Substituting back for dx/du, we have:
= ∫ (sinx - (sinx + cosx)(e^(sinx + cosx))) / u^2 * e^(sinx)cosx * 1 / (e^(sinx)cosx) dx/du
= ∫ (sinx - (sinx + cosx)(e^(sinx + cosx))) / u^2 * du
Expanding the expression, we get:
= ∫ (sinx - sinx * e^(sinx + cosx) - cosx * e^(sinx + cosx)) / u^2 * du
= ∫ (sinx - sinx * e^(sinx + cosx))/(e^(sinx))^2 * du - ∫ cosx/e^(sinx) * du
Now, we can break down the integral into two separate integrals:
∫ (sinx - sinx * e^(sinx + cosx))/(e^(sinx))^2 * du - ∫ cosx/e^(sinx) * du
= ∫ sinx/u^2 * du - ∫ cosx/u * du
Integrating the first term, we have:
= -cosx/u - ∫ -cosx * (d(1/u))/dx
Notice that -cosx * (d(1/u))/dx is the derivative with respect to x of (sinx/u). So, we can rewrite it as:
= -cosx/u - -(sinx/u) + ∫ sinx/u^2 * du
= -cosx/u + sinx/u + ∫ sinx/u^2 * du
Now, we can simplify the integral:
∫ sinx/u^2 * du = ∫ (sindx/u)(du/u)
= - (1/u)(cosx) + ∫ (1/u)(d(sinx))/dx dx
Integration of (1/u)(d(sinx))/dx is straightforward:
= - (1/u)(cosx) + ∫ (1/u)cosx dx
= - (1/u)(cosx) + (1/u)sinx + C, where C is the constant of integration
Finally, substituting back u = e^(sinx), we get the complete solution for the integral:
= - (1/e^(sinx))(cosx) + (1/e^(sinx))sinx + C
= - cosx / e^(sinx) + sinx / e^(sinx) + C
Therefore, the integral of the given expression is:
∫ (e^(cosx)sinx - (sinx + cosx)e^(sinx+cosx)) / (e^(sin2x) - 2e^(sinx) + 1) dx = - cosx / e^(sinx) + sinx / e^(sinx) + C, where C is the constant of integration.
Oops. That should have been
ecosxsinx - (sinx+cosx)esinx+cosx
-----------------------------------------
e2sinx-2esinx+1