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intenzxboi
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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?
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).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
Mark got in just ahead of me but I think my suggestion for integrating it is easier!u= sin x
du= cos x
so i get
1/7 ln sinx + C
then plug pi/2 minus pi/6?
Yes, I agree. It just goes to show that "haste makes waste."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!
The integral of 1/7ln(sin x) between pi/2 and pi/6 is approximately -0.054.
To evaluate the integral of 1/7ln(sin x) between pi/2 and pi/6, you can use the substitution method by letting u = sin x and du = cos x dx. This will transform the integral into ∫1/7ln(u) du, which can be evaluated using basic integration rules.
The integral of 1/7ln(sin x) between pi/2 and pi/6 is a definite integral, as it has specific limits of integration.
Yes, the integral of 1/7ln(sin x) between pi/2 and pi/6 can be solved using a calculator by first converting it into a simpler form, such as ∫1/7ln(u) du, and then using the integration function on the calculator to find the solution.
No, the integral of 1/7ln(sin x) between pi/2 and pi/6 can be evaluated using basic integration techniques such as substitution and integration by parts. However, it may require some algebraic manipulation to simplify the integrand before integrating.