Evaluate Integral: 1/7ln(sin x) b/w pi/2 & pi/6

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Homework Statement



between pi/2 and pi/6[tex]\int[/tex] (cos x) / (7 + sin x)


move 1/7 to the out side

1/7 [tex]\int[/tex] 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?
 
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You can't do that.

Is [tex]\frac{1}{7 + 3} = 1/7 * \frac{1}{3}[/tex]
?

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

Homework Statement



between pi/2 and pi/6[tex]\int[/tex] (cos x) / (7 + sin x)


move 1/7 to the out side

1/7 [tex]\int[/tex] 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!
 
Last edited by a moderator:
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
 
HallsofIvy said:
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."