## bettyboop8904 2 years ago Ok I have another integral problem. Equation below

1. bettyboop8904

$\int\limits_{}^{}\arctan(4t)dt$

2. khoala4pham

Well, the full solution is quite long. The general solution is that you have to use integration by parts: $\int\limits_{}^{}udv = uv - \int\limits_{}^{}vdu$ Here, if you let u = arctan 4t, dv = dt you obtain $du=\frac{ 4 }{16+t^{2}}dt$ and v = t. Then, you substitute back into the equation and solve. The integral of vdu should (if I'm doing my mental calculation correctly) be a u-sub for u as the denominator.

3. bettyboop8904

why wouldn't you make just arctan u?

4. khoala4pham

Because there is no integral formula for arctanu. We know the derivative of arctan u but the integral of arctan u must be done with integration by parts.

5. bettyboop8904

right but what i mean can we set u=arctan and dv=4t? or does the dt always have to be with one of the items you assign?

6. bettyboop8904

and wouldn't du above be $\frac{ 4 }{ 1+16t ^{2} }$

7. khoala4pham

Well the thing is that u, v, du, and dv must all be functions. arctan by itself is not a function. What are you evaluating arctan at? You are right though. dv must be attached to dt. The purpose of integration by parts is to obtain all values u, v, du, and dv. if you had let u = dt, then how would you solve for u? If you integrate both sides, what are you integrating u with respect with? If you differentiate u, the derivative of dt has no meaning. And the answer is no to your question, we have u = arctan4t. If you treat the derivative as a fraction, you get $\frac{ du }{ dt } = \frac{ 1 }{ 16+t ^{2} }$, then you must "multiply" both sides by dt. I must make the obligatory statement that du/dt is NOT NOT NOT NOT NOT a fraction. I will not be able to explain to you why this works without a treatment in differential forms which will confuse more than clarify but trust that it works. You can IMAGINE it as a fraction.

8. khoala4pham

Perhaps your confusion about integration by parts arises from what it all means. The notation is highly obscure but you need to re-examine what it means. Ultimately, all the equations of u, v, and their derivatives are dependent upon the variable t. However, with the right choices, we can use u, v, du, and dv (remember! they are dependent upon t!!) to rearrange the integral into something much easier to solve. If you look at my original work, you will see I made the choice u = arctan 4t and dv = dt. u here is obviously a function of t. For dv = dt, if you integrate both sides you get v = t, another function of t. Ultimately, evaluating the right hand side of the equation will amount to $t*\arctan(4t) - \int\limits_{}^{}\frac{ 4t }{ 16+t ^{2} }dt$ Wow. All these expressions are in terms of t. Now you just have to evaluate the integral with another proper method: u-substitution. Integration by parts is merely a way to rearrange integrals. It's the analogous integral version of the product rule of derivatives.

9. bettyboop8904

ok i still don't get how you're getting that = ( i know the derivative of arctan is 1/1+x^2 what is is the derivative of arctan(f(x))? isnt is 1/1+(f(x))^2 * f'(x)??

10. khoala4pham

Your notation is slightly ambiguous: the derivative of arctan(f(x)) with respect to x is: (f'(x))/(1+f(x)^2). Here, f(t) = 4t. f'(t) = 4.

11. khoala4pham

Oh I see what you're saying. My bad. I type slower than I think. It is 1 + 16t^2.

12. bettyboop8904

no thats how i meant i just put it at the end. usually i use the equation thingy but it takes me forever to post then but i meant for f'(x) to be in the numerator

13. bettyboop8904

lol

14. khoala4pham

If I'm understanding you correctly, you're absolutely right. I was in a rush to post a reply. The proper evaluation is $t*\arctan(4t) - \int\limits_{}^{}\frac{ 4t }{ 1 + 16t ^{2} }dt$

15. bettyboop8904

ok now what would i do from there with the next integral? i'm sorry I have so much trouble with these = (

16. bettyboop8904

would i use the 1/a srctan(x/a) + c?

17. bettyboop8904

*arctan

18. bettyboop8904

or let u = 4tdt? = (

19. khoala4pham

Now you do a u substitution. Why? because you noticed that the denominator is a power of 2 while the numerator is a power of 1. If we let u = 1 + 16t^2, we get du = 32 t dt. That means the integral is $\frac{ 1 }{ 8 }\int\limits_{}^{}\frac{ du }{ u }$ which is obviously the ln|u| + c. Remember two things, first it's the natural log of the absolute value of u, and second, you are subtracting this integral. After that, you can re-substitute u = 1 + 16t^2 into the natural log and voila, you have the answer.

20. khoala4pham

To your first question, no you cannot reduce it to the arctan formula. The arctan formula only works if your numerator is 1. To your second question, are you trying to do another integration by parts? That would be unwise since you can do a u-sub.

21. bettyboop8904

you make it sound so easy lol i know how to do almost all of the parts when dealing with integrating and derivatives but i just can't seem to put it all together = (((

22. bettyboop8904

ok are you up to help me with another one by chance?

23. bettyboop8904

@khoala4pham

24. khoala4pham

No problem @bettyboop8904