Integral from -pi/3 to 0 of 3sin^3(x)dx ?

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In summary, the conversation discusses using integration by parts to solve the integral ∫3sec3x dx. After the initial attempt, it is realized that a different approach is needed and the integral can be rewritten as ∫tan2x secx dx. By further simplifying, the integral can be solved algebraically to find the final answer.
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Lo.Lee.Ta.
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∫from [itex]-\pi[/itex]/3 to 0 of 3sec3xdx

Integration by parts...

3∫sec2x*secxdx

dv= 3sec2x
v= 3tanx

u= secx
du= secxtanx

Formula: uv - ∫vdu

secx*3tanx - ∫3tanx*secxtanxdx

Integration by parts again...

u= 3tanx
du= 3sec2x

dv= sextanx
v= secx

3tanx*secx - ∫secx*3sec2x

Altogether:

∫3sec3x = 3secxtanx - 3secxtanx - ∫3sec3x

*Add ∫3sec3x on both sides*

∫2*3sec3x = 0 !

It seems that the two 3secxtanx's would cancel! :/

∫6sec3x = 0

This is like the same thing we started out with! :confused: ugh, man. I don't really know what to do now.
I know this is a definite integral, so we should substitute -pi/3 and 0 in the end, but we still have an integral! #=_=

...This definitely does not seem right! :(
But I don't really see what's wrong with my work.
Please help! Thank you so much! :D
 
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Lo.Lee.Ta. said:
∫from [itex]-\pi[/itex]/3 to 0 of 3sec3xdx

Integration by parts...

3∫sec2x*secxdx

dv= 3sec2x
v= 3tanx

u= secx
du= secxtanx

Formula: uv - ∫vdu

secx*3tanx - ∫3tanx*secxtanxdx

Integration by parts again...
That's not the way to go. Your integral above can be written as
## \int tan^2(x) sec(x) dx##
## = \int (sec^2(x) - 1) sec(x) dx ##
This will give you another sec3 integral and one that's somewhat easier. You should be able to solve algebraically for the integral you're looking for.
I am not including the factor of 3 from the first integral. It would have been better to bring it out of the integral right away and just concentrate on ∫sec3(x) dx.
Lo.Lee.Ta. said:
u= 3tanx
du= 3sec2x

dv= sextanx
v= secx

3tanx*secx - ∫secx*3sec2x

Altogether:

∫3sec3x = 3secxtanx - 3secxtanx - ∫3sec3x

*Add ∫3sec3x on both sides*

∫2*3sec3x = 0 !

It seems that the two 3secxtanx's would cancel! :/

∫6sec3x = 0

This is like the same thing we started out with! :confused: ugh, man. I don't really know what to do now.
I know this is a definite integral, so we should substitute -pi/3 and 0 in the end, but we still have an integral! #=_=

...This definitely does not seem right! :(
But I don't really see what's wrong with my work.
Please help! Thank you so much! :D
 

FAQ: Integral from -pi/3 to 0 of 3sin^3(x)dx ?

1. What is the value of the integral from -pi/3 to 0 of 3sin^3(x)dx?

The value of the integral is approximately 0.5.

2. How do you solve the integral from -pi/3 to 0 of 3sin^3(x)dx?

To solve this integral, you can use the substitution method by letting u = sin(x) and du = cos(x)dx. This will transform the integral into -3/4 times the integral of u^3 from -1/2 to 0, which can then be solved using the power rule for integration.

3. Why is the upper limit of the integral -pi/3 and the lower limit 0?

The limits of integration are determined by the given problem or application. In this case, the integral is being evaluated over the interval from -pi/3 to 0 because this is the range of values for which the function 3sin^3(x) is defined and the area under the curve is of interest.

4. What does the integral from -pi/3 to 0 of 3sin^3(x)dx represent?

This integral represents the area under the curve of the function 3sin^3(x) over the interval from -pi/3 to 0. It can also be interpreted as the net change in the position of an object, given its velocity is 3sin^3(x).

5. Can the integral from -pi/3 to 0 of 3sin^3(x)dx be evaluated using other methods?

Yes, there are other methods that can be used to solve this integral, such as integration by parts or trigonometric identities. However, the substitution method is the most straightforward and efficient method for this particular integral.

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