Thread: quick help with 3 integrals

1. quick help with 3 integrals

NOTE: I use ( as the beggining intergral sign... so ( x^2 dx is the integral of x^2.

1.I have two integrals. (tan x sec^2 x dx and (sin x / cos^3 x dx
-after looking at them I realize they are the same thing...
-then why do i get different answers?
1) u =tan x du=sec^2 x dx ((u)du = u^2 /2. Answer is (tan^2 x/2) + C
2) u =cosx du=-sinx dx -du=sinx dx (-du/(u)^3= (u)^-2/2. Answer is (sec^2 x)/2 + C
What am I doing wrong?

2. (tan(3x) dx
- I got the answer -1/3 ln cos(3x) +C. I was wondering if I should an absolute value around COS 3x or not. Should I?

3.( x(2+x)^.5 dx
-how do i do this one?
- this is how far i got. u= 2+x du=1 ( du(x) (u)^.5 I am lost now.

-Thanks for the help in advance

2. Originally Posted by dandaman
NOTE: I use ( as the beggining intergral sign... so ( x^2 dx is the integral of x^2.

1.I have two integrals. (tan x sec^2 x dx and (sin x / cos^3 x dx
-after looking at them I realize they are the same thing...
-then why do i get different answers?
1) u =tan x du=sec^2 x dx ((u)du = u^2 /2. Answer is (tan^2 x/2) + C
2) u =cosx du=-sinx dx -du=sinx dx (-du/(u)^3= (u)^-2/2. Answer is (sec^2 x)/2 + C
What am I doing wrong?

2. (tan(3x) dx
- I got the answer -1/3 ln cos(3x) +C. I was wondering if I should an absolute value around COS 3x or not. Should I?

3.( x(2+x)^.5 dx
-how do i do this one?
- this is how far i got. u= 2+x du=1 ( du(x) (u)^.5 I am lost now.

-Thanks for the help in advance

$\int \frac{\sin x}{\cos^3x}dx=\int -u^{-3}du=\frac{1}{2}u^{-2}=\frac{1}{2 \cos^2 x}=\frac{\sec^2x}{2}+C$ $=\frac{\tan^2x+1}{2}+C=\frac{\tan^2x}{2}+\left(\fr ac{1}{2}+C\right)$ <--It is just a different C.

3. k thanks... any idea for the others?

4. Originally Posted by dandaman
NOTE: I use ( as the beggining intergral sign... so ( x^2 dx is the integral of x^2.

1.I have two integrals. (tan x sec^2 x dx and (sin x / cos^3 x dx
-after looking at them I realize they are the same thing...
-then why do i get different answers?
1) u =tan x du=sec^2 x dx ((u)du = u^2 /2. Answer is (tan^2 x/2) + C
2) u =cosx du=-sinx dx -du=sinx dx (-du/(u)^3= (u)^-2/2. Answer is (sec^2 x)/2 + C
What am I doing wrong?

2. (tan(3x) dx
- I got the answer -1/3 ln cos(3x) +C. I was wondering if I should an absolute value around COS 3x or not. Should I?

3.( x(2+x)^.5 dx
-how do i do this one?
- this is how far i got. u= 2+x du=1 ( du(x) (u)^.5 I am lost now.

-Thanks for the help in advance
Originally Posted by dandaman
k thanks... any idea for the others?
Sorry I was taking off yesterday and did not have time to come back until now.

For #2, I believe you should have absolute value since you are talking about $\ln$.
For #3, with your choice of u,
since $u=2+x$, $du=dx$. Thus $x=u-2$
So $\int x(2+x)^{\frac{1}{2}}dx=\int (u-2)u^{\frac{1}{2}}du$ $=\int u^{\frac{3}{2}}-2u^{\frac{1}{2}}du=...$.
I believe you can take it from here.