# integration (beginner)

• May 12th 2009, 12:06 PM
Bernice
integration (beginner)
I've just started to learn about integration and can't even solve these examples:

a)$\displaystyle \int (x^2-\frac{1}{2\sqrt{x}})^2\, dx$

b)$\displaystyle \int (sin\frac{x}{2}+cos\frac{x}{2})^2\, dx$

Could someone show me how to do it?
• May 12th 2009, 12:22 PM
Spec
a)$\displaystyle \int \left(x^2-\frac{1}{2\sqrt{x}}\right)^2\, dx=\int \left(x^4-x^{3/2}+\frac{1}{4x}\right)dx$

b)$\displaystyle \int (sin\frac{x}{2}+cos\frac{x}{2})^2\, dx=\int \left(\sin^2 \frac{x}{2} +2\cos \frac{x}{2}\sin \frac{x}{2} +\cos^2 \frac{x}{2}\right)dx$

a) Should be easy to solve from the last step.

b) Use the trigonometric identitities: $\displaystyle 2\cos^2 \frac{x}{2} = 1+\cos x$, $\displaystyle 2\sin^2 \frac{x}{2} = 1-\cos x$ and note that $\displaystyle 2\cos \frac{x}{2}\sin \frac{x}{2}=\frac{d}{dx}\left(2\sin^2\frac{x}{2} \right)$
• May 12th 2009, 12:53 PM
I would maybe guess that you're having trouble with figuring out how to integrate them rather than expanding it so, continuing from Specs post.

You can separate the integral in a) to be...
$\displaystyle \int x^4 dx - \int x^{3/2} dx + \int \frac{1}{4x} dx$

And then using the fact that $\displaystyle \int x^n dx = \frac{x^{n+1}}{n+1}$, you should be able to solve these.

For part b) use $\displaystyle \int \sin(x) dx = -\cos(x)$ and $\displaystyle \int \cos(x) dx = \sin(x)$ after you've used the identities given in Specs post.
• May 12th 2009, 12:59 PM
Bernice
Thank you!
So a)$\displaystyle \int \left(x^2-\frac{1}{2\sqrt{x}}\right)^2\, dx=\int \left(x^4-x^{3/2}+\frac{1}{4x}\right)dx=\frac{x^5}{5}-\frac{2\sqrt{x^5}}{5}+\frac{1}{4}ln|x|+C$$\displaystyle =\frac{x^5}{5}-\frac{2x^2\sqrt{x}}{5}+\frac{1}{4}ln|x|+C$
• May 12th 2009, 01:17 PM
Bernice
And for b) I got: $\displaystyle x-cosx+C$
• May 12th 2009, 01:25 PM
Spec
EDIT: Nevermind, didn't notice that you rewrote the answer using the identity above.
• May 12th 2009, 01:45 PM
Bernice
I've another 2 examples and I'm not sure how to work out them:

c)$\displaystyle \int cos(\frac{3}{7}x+2)\, dx$

d)$\displaystyle \int \frac{2}{3x+5}\, dx$

For c) I tryed:

$\displaystyle \int cos(\frac{3}{7}x+2)\, dx =*$

$\displaystyle \frac{3}{7}x+2=t$

$\displaystyle d(\frac{3}{7}x+2)=dt$

$\displaystyle \frac{3}{7}dx=dt$

$\displaystyle dx=\frac{7dt}{3}$

$\displaystyle *=\int cos \ t\frac{7dt}{3}=\frac{7}{3}\int cos \ t \, dt =\frac {7}{3} sin \ t+C=\frac{7}{3}sin(\frac{3}{7}x+2)+C$

Does it look correct?
• May 12th 2009, 02:00 PM
Spec
c) looks correct.

For d), use the fact that $\displaystyle \frac{2}{3} \cdot \frac{d}{dx}\left( \ln(3x+5)\right)=\frac{2}{3} \cdot \frac{3}{3x+5}=\frac{2}{3x+5}$
• May 12th 2009, 03:37 PM
Bernice
One last question in this thread:

e) $\displaystyle \int e^{2-0.2x}\,dx$

Is the answer $\displaystyle -5e^{2-0.2x}+C$ ?
• May 12th 2009, 03:41 PM
derfleurer
$\displaystyle e^{2 - 0.2x} = e^2e^{-0.2x}$

Where $\displaystyle e^2$ is just a constant. Use a u substitution.

$\displaystyle e^2 \int e^{-0.2x}dx = \frac{-e^2}{0.2}e^{-0.2x} = \frac{-e^{2-0.2x}}{0.2}$