evaluate integral

[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???

[Mark44]

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)).

[HallsofIvy]

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
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.

u= sin x
du= cos x

so i get
1/7 ln sinx + C

then plug pi/2 minus pi/6???
Mark got in just ahead of me but I think my suggestion for integrating it is easier!

[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

[Mark44]

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