# Indefinite Integral with integration by parts

• Physics-UG
In summary, the student is attempting to solve a homework problem from calculus II, but seems to be stuck. They have substituted the correct substitutions, but are not using the exponential and trigonometric functions together correctly.
Physics-UG

## Homework Statement

Evaluate ∫ecos2θ dθ

## Homework Equations

Integration by parts formula
∫udv = uv -∫vdu

## The Attempt at a Solution

So in calc II we just started integration by parts and I'm doing one of the assignment problems. I know I need to do the integration by parts twice, but I've hit a loop, or so it seems, which I know isn't right. Maybe someone can see my faults?

I set:
u = cos(2θ) v = -e
du = -2sin(2θ) dθ dv = e

So by the formula I got:
= [cos(2θ)] [-e] -2∫ [-e] [sin(2θ) dθ]

Here I used the second integration by parts:

u = sin2θ v = e
du = 2cos(2θ) dθ dv = -e

Solving by the formula again:
= [sin(2θ)] [e] -2∫ [e] [cos(2θ) dθ]

I'm not too sure where I've made my algebraic error, or if I'm on the right track and this won't just put my into a loop giving the same equation above? First question I've posted, so hopefully it follows the format ok, if it doesn't, chime in and let me know so I can fix it properly. Thanks in advance everyone!

Your substitutions are correct, but you're stopping a little short. The trick with integration by parts composed of an exponential function and a trigonometric function is to add to both sides the first integral you find that is a multiple of your original integral. From there, it should be obvious where to go next.

phion said:
Your substitutions are correct, but you're stopping a little short. The trick with integration by parts composed of an exponential function and a trigonometric function is to add to both sides the first integral you find that is a multiple of your original integral. From there, it should be obvious where to go next.
Ohh, hat's definitely what I'm missing! Thanks a bunch for the help!

## What is an indefinite integral with integration by parts?

An indefinite integral with integration by parts is a method for finding the antiderivative of a function by using the product rule of differentiation. It is used when the function to be integrated is a product of two functions.

## How do you solve an indefinite integral with integration by parts?

To solve an indefinite integral with integration by parts, you need to follow these steps:

1. Choose one function as u and the other as dv.
2. Find the derivative of u (du) and the antiderivative of dv (v).
3. Use the formula ∫u dv = uv - ∫v du to solve the integral.

## What are the conditions for using integration by parts?

The conditions for using integration by parts are:

• The integral must be in the form of ∫u dv.
• The function u must be differentiable.
• The function dv must be integrable.

## Can integration by parts be used for definite integrals?

Yes, integration by parts can be used for definite integrals. In this case, the formula becomes ∫a^b u(x) dv(x) = [u(x)v(x)]^b_a - ∫a^b v(x) du(x), where a and b are the limits of integration.

## What are some common applications of integration by parts?

Integration by parts is commonly used in calculus and physics to solve various problems, such as finding the area under a curve, calculating work done by a force, and finding the center of mass of an object. It is also used in probability and statistics to find the expected value of a random variable.

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