# Integrate cos(lnx)dx - LIPTE Rule Help

• Cudi1
In summary, the student is trying to solve an integration by parts problem and is getting stuck. They are using the LIPTE rule and are confused by the cosdx term. After asking for help, they are able to solve the problem by substitution.
Cudi1

let u=lnx
du=1/x*dx
dv=cosdx
v=-sin

## Homework Equations

Now I am confused as I am getting nowhere with this substiution, i learned the LIPTE rule but its quite confusing, i have a function within a function

## The Attempt at a Solution

for integration by parts to work, i would need two differentiable functions, but the cosdx is not differentiable would it need to be cosxdx for it to be differentiated?

Is the integral:

$$\int{\cos{(\ln{(x)})} \, dx}$$

If it is, make the substitution:

$$t = \ln{(x)} \Rightarrow x = e^{t}$$

and substitute everywhere. The integral that you will get can be integrated by using integration by parts twice, or, if you know complex numbers, by representing the trigonometric function through the complex exponential.

it is just cos(lnx)dx

then proceed as I told you.

ok I am getting an integral of the form : coste^tdt. is this correct?

yes. proceed by integration by parts or using complex exponentials.

ok thank you for the help, quick question why do we have to let t=lnx?

We have a composite function $\cos{(\ln{x})}$ of two elementary functions (trigonometric and logarithmic). As a combination, it does not have an immediate table integral. But, the method of substitution, which is nothing but inverting the chain rule for derivatives of composite functions, works exactly for such compound functions.

thank you, lastly i end up with 1/2(e^tcost-e^tsint), do i sub back , so that x=e^t and t=lnx
which leaves me with : 1/2x(cos(lnx)-sin(lnx)+c?

I think the sign in front of the sine should be + and the result should be:

$$\frac{1}{2} x \left[\cos{\left(\ln{(x)}\right)} + \sin{\left(\ln{(x)}\right)}\right]+ C$$

yes, made a slight mistake thank you very much for the help, i tried doing it another way by letting u=cos(lnx) and dv=dx and i arrived at the same answer

yes, the x's will cancel.

## 1. What is the LIPTE rule?

The LIPTE rule is a mnemonic device used to remember the steps for integrating by parts. LIPTE stands for: Logarithmic, Inverse trigonometric, Polynomial, Trigonometric, and Exponential. It helps identify which function should be chosen as u and which as dv when using the integration by parts method.

## 2. Can the LIPTE rule be used for all integrals?

No, the LIPTE rule is not applicable to all integrals. It is most commonly used for integrals involving products of functions, where the product rule cannot be easily applied. It is also useful for integrals involving logarithmic, inverse trigonometric, polynomial, trigonometric, and exponential functions.

## 3. How do you use the LIPTE rule?

To use the LIPTE rule, follow these steps: 1) Identify the functions in the integral that fall under the LIPTE categories. 2) Choose one as u and the other as dv. 3) Take the derivative of u and the antiderivative of dv. 4) Plug these values into the integration by parts formula: ∫udv = uv - ∫vdu. 5) Simplify the resulting integral and solve for the original integral.

## 4. Can the LIPTE rule be applied to this specific integral: ∫cos(lnx)dx?

Yes, the LIPTE rule can be applied to this integral. In this case, u = ln(x) and dv = cos(x)dx. The derivative of u is 1/x and the antiderivative of dv is sin(x). Plugging these values into the integration by parts formula results in ∫cos(lnx)dx = ln(x)sin(x) - ∫(1/x)sin(x)dx. This integral can then be solved using the substitution method.

## 5. Are there any other methods for integrating cos(lnx)dx?

Yes, there are other methods for integrating this function, such as using the power rule or substitution method. However, the LIPTE rule is often the most efficient and straightforward method for this specific integral.

• Calculus and Beyond Homework Help
Replies
15
Views
775
• Calculus and Beyond Homework Help
Replies
4
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
482
• Calculus and Beyond Homework Help
Replies
6
Views
721
• Calculus and Beyond Homework Help
Replies
22
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
943
• Calculus and Beyond Homework Help
Replies
1
Views
850
• Calculus and Beyond Homework Help
Replies
10
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
2K
• Calculus and Beyond Homework Help
Replies
8
Views
754