Differentiation under integral sign

In summary, the conversation discusses solving indefinite integrals using differentiation under the integral sign and integration by parts. The formulas for $\displaystyle\int \cos(tx)\,dx$ and $\displaystyle\int \sin(tx)\,dx$ are given, and it is mentioned that setting $t=1$ will result in the correct answers. The formula for $\displaystyle\int x^4\sin{(x)}dx$ is derived using this method, and it is stated that similar problems will occur for other values of n. The conversation ends with a suggestion to differentiate both sides of the given formulas with respect to t to get all the necessary formulas.
  • #1
WMDhamnekar
MHB
376
28
Hello,

How to find formulas for these$\displaystyle\int x^n\sin(x)\, dx, \displaystyle\int x^n\cos(x)\, dx,$ indefinite integrals when $n=1,2,3,4$ using differentiation under the integral sign starting with the formulas

$$\displaystyle\int \cos(tx)\,dx = \frac{\sin(tx)}{t}, \displaystyle\int \sin(tx)\,dx= -\frac{\cos(tx)}{t}$$ for $t > 0.$

I don't have any idea to solve these indefinite integrals except to solve them recursively using integration by parts.
 
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  • #2
You are required to use "differentiation" under the integral? These can be done directly using "integration by parts".
 
  • #3
See what happens when you differentiate both sides of those relations with respect to $t$ and then set $t=1$.
 
  • #4
MountEvariste said:
See what happens when you differentiate both sides of those relations with respect to $t$ and then set $t=1$.

Hello,
I solved the two indefinate integrals using differentiation under integral sign as well as using integration by parts. Under these both method i got the correct answers. For example if we put n=4,the integral to be solved becomes $\displaystyle\int x^4\sin{(tx)}dx =\frac{(4x^3t^3-24tx)*\sin{(tx)}+(-x^4t^4+12x^2t^2-24)*\cos{(tx)}}{t^5}$

Now if we put t=1, the answer will become $(4x^3-24x)*\sin{(x)}+(-x^4+12x^2-24)*\cos{(x)}$

Now how to derive the formula for $\displaystyle\int x^4\sin{(x)}dx?$

In other cases where n=1,2,3, similar problem will persist.
 
  • #5
Dhamnekar Winod said:
In other cases where n=1,2,3, similar problem will persist.
Perhaps you've figured it out by now, but I don't see any differentiation under integral sign.

$$\displaystyle\int \cos(tx)\,dx = \frac{\sin(tx)}{t}, \displaystyle\int \sin(tx)\,dx= -\frac{\cos(tx)}{t}$$ for $t > 0.$
In these two equalities, differentiate both the RHS and the LHS with respect to $t$. Adjust for constant. Do it again, and again. You'll get all of them.
 

Related to Differentiation under integral sign

1. What is the differentiation under integral sign?

Differentiation under integral sign is a mathematical technique used to find the derivative of a function that contains an integral as part of its expression. It allows us to differentiate the function with respect to a variable that is also present in the limits of the integral.

2. Why is differentiation under integral sign important?

This technique is important because it allows us to solve complex integrals that cannot be evaluated using traditional methods. It also has practical applications in physics, engineering, and other fields where integrals are commonly used.

3. How does differentiation under integral sign work?

The process involves swapping the integral and the differentiation operators, then evaluating the resulting integral. This is done using the Fundamental Theorem of Calculus, which states that the derivative of an integral is equal to the integrand evaluated at the upper limit of integration.

4. What are some common mistakes to avoid when using differentiation under integral sign?

Some common mistakes include forgetting to swap the integral and differentiation operators, incorrectly evaluating the resulting integral, and not paying attention to the limits of integration. It is important to carefully follow the steps and double check your work to avoid these errors.

5. Can differentiation under integral sign be used for all types of integrals?

No, this technique can only be used for integrals that have a variable present in both the integrand and the limits of integration. It cannot be used for integrals with constant limits or where the variable is only present in the integrand.

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