Math Problem Statement

solve the equations

Solution

Let's solve each of the given integration problems one by one:

a) lnxxdx\int \frac{\ln x}{x} \, dx

To solve this, we can use the substitution u=lnxu = \ln x, then du=1xdxdu = \frac{1}{x} \, dx.

lnxxdx=udu=u22+C=(lnx)22+C\int \frac{\ln x}{x} \, dx = \int u \, du = \frac{u^2}{2} + C = \frac{(\ln x)^2}{2} + C

So, the solution is:

(lnx)22+C\boxed{\frac{(\ln x)^2}{2} + C}

b) exsin(ex)dx\int e^x \sin(e^x) \, dx

Let u=exu = e^x, then du=exdxdu = e^x \, dx. This transforms the integral into:

exsin(ex)dx=sin(u)du\int e^x \sin(e^x) \, dx = \int \sin(u) \, du

The integral of sin(u)\sin(u) is cos(u)-\cos(u), so:

sin(u)du=cos(u)+C=cos(ex)+C\int \sin(u) \, du = -\cos(u) + C = -\cos(e^x) + C

Thus, the solution is:

cos(ex)+C\boxed{-\cos(e^x) + C}

c) x3x2+11dx\int x^3 \sqrt{x^2 + 11} \, dx

Let u=x2+11u = x^2 + 11. Then, du=2xdxdu = 2x \, dx, or du2=xdx\frac{du}{2} = x \, dx.

We rewrite x3x2+11x^3 \sqrt{x^2 + 11} as x2xx2+11x^2 \cdot x \sqrt{x^2 + 11}. Then, x2=u11x^2 = u - 11:

x3x2+11dx=(u11)udu2\int x^3 \sqrt{x^2 + 11} \, dx = \int (u - 11) \sqrt{u} \cdot \frac{du}{2}

This simplifies to:

12(u11)u1/2du=12(u3/211u1/2)du\frac{1}{2} \int (u - 11) u^{1/2} \, du = \frac{1}{2} \int (u^{3/2} - 11u^{1/2}) \, du

Integrate each term separately:

12(25u5/21123u3/2)+C\frac{1}{2} \left( \frac{2}{5} u^{5/2} - 11 \cdot \frac{2}{3} u^{3/2} \right) + C

Simplifying this:

15u5/2113u3/2+C\frac{1}{5} u^{5/2} - \frac{11}{3} u^{3/2} + C

Substitute u=x2+11u = x^2 + 11:

15(x2+11)5/2113(x2+11)3/2+C\boxed{\frac{1}{5} (x^2 + 11)^{5/2} - \frac{11}{3} (x^2 + 11)^{3/2} + C}

d) elnxcos2xdx\int \frac{e^{\ln x}}{\cos^2 x} \, dx

First, notice that elnx=xe^{\ln x} = x:

xcos2xdx=xsec2xdx\int \frac{x}{\cos^2 x} \, dx = \int x \sec^2 x \, dx

We can use integration by parts for this integral. Let u=xu = x and dv=sec2xdxdv = \sec^2 x \, dx. Then, du=dxdu = dx and v=tanxv = \tan x:

xsec2xdx=xtanxtanxdx\int x \sec^2 x \, dx = x \tan x - \int \tan x \, dx

The integral of tanx\tan x is lncosx-\ln |\cos x|:

xtanx+lncosx+Cx \tan x + \ln |\cos x| + C

Thus, the solution is:

xtanx+lncosx+C\boxed{x \tan x + \ln |\cos x| + C}

If you have any questions or need further details on any of these integrations, feel free to ask! Here are some related questions to explore the topic further:

  1. What is the general method for integrating functions of the form lnxx\frac{\ln x}{x}?
  2. How do you identify appropriate substitutions for integrals involving composite functions?
  3. What are the common techniques used in integration by parts?
  4. Can you derive the formula for integrating exsin(ex)e^x \sin(e^x) without substitution?
  5. How do you handle integrals involving square roots of quadratic expressions?
  6. What are some examples of integrals that require a combination of substitution and parts?
  7. How does the integration of trigonometric functions differ from polynomial functions?
  8. What are some common mistakes to avoid when performing integration by parts?

Tip: Always verify your integration results by differentiating your answer to see if you obtain the original integrand. This ensures the accuracy of your solution.