# Evaluate integral

1. Feb 6, 2009

### intenzxboi

1. The problem statement, all variables and given/known data

between pi/2 and pi/6$$\int$$ (cos x) / (7 + sin x)

move 1/7 to the out side

1/7 $$\int$$ cos x / sin x

u= sin x
du= cos x

so i get
1/7 ln sinx + C

then plug pi/2 minus pi/6???

2. Feb 6, 2009

### Staff: Mentor

You can't do that.

Is $$\frac{1}{7 + 3} = 1/7 * \frac{1}{3}$$
?

Instead, you might try multiplying by (7 - sin(x))/(7 - sin(x)).

3. Feb 6, 2009

### HallsofIvy

Staff Emeritus
If you are taking calculus, your algebra should be better than that! cos(x)/(7+ sin(x)) is NOT (1/7) cos(x)/sin(x).

Instead use the substitution u= 7+ sin x.

Mark got in just ahead of me but I think my suggestion for integrating it is easier!

Last edited: Feb 6, 2009
4. Feb 6, 2009

### intenzxboi

k got it
i have taken any math for 2 year and i am now taking cal 2 so yea... but i solved it thanks

5. Feb 6, 2009

### Staff: Mentor

Yes, I agree. It just goes to show that "haste makes waste."