Integration by parts help just the beginning part for this one

In summary, the conversation discusses the use of substitution to solve for the integral from pi to 0 of (sin(3t)dt)^4. The person mentions struggling with getting the correct coefficient when using the u/du substitution method and wonders if they are using it improperly. They also mention not needing to change the limits when using substitution.
  • #1
camboguy
36
0

Homework Statement


intergral from pi to 0.
of
(sin(3t)dt)^4






The Attempt at a Solution



okay so i know how to do this but when i tried substitution putting 3t=u and (1/3)du= dt i always came with the the wrong coefficient at the end with the answer and so i would multiply it wrong. but then when i do not use u/du substitution everything works well? why is that? what is the difference? and I am using u/du substitution improperly or something?
 
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  • #2
Are you correctly changing the limits?
 
  • #3
i didn't need to change it because i just plugged back in the variables u with 3t. just wondering it should be correct even if i substitute right?
 

FAQ: Integration by parts help just the beginning part for this one

1. What is integration by parts?

Integration by parts is a method used in calculus to evaluate integrals of products of functions. It is based on the product rule for differentiation and allows for the integration of functions that cannot be easily integrated using other methods.

2. What is the formula for integration by parts?

The formula for integration by parts is ∫ u dv = u*v - ∫ v du, where u and v are functions and du and dv are their respective differentials. This formula is derived from the product rule for differentiation.

3. When should I use integration by parts?

Integration by parts is useful when the integral of a product of functions cannot be easily evaluated using other methods, such as substitution or trigonometric identities. It is also helpful when the integrand contains a polynomial and an exponential or logarithmic function.

4. How do I choose which function to use for u and dv?

When using integration by parts, it is important to choose u and dv in a way that simplifies the integral or reduces it to a known integral. A common method is to choose u as the function that is most easily differentiated and dv as the function that is most easily integrated.

5. Can integration by parts be used for definite integrals?

Yes, integration by parts can be used for definite integrals. After applying the integration by parts formula, the resulting integral can be evaluated using the limits of integration. However, it may be necessary to use integration by parts multiple times for more complicated definite integrals.

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