Question

I'm confused as to how my textbook got the answer for the following problem. Can someone walk me through how they get the answer? I'm so close, but my answer isn't quite there and I'm not sure on the reasoning for how to get to the book's answer.

Any and all help is greatly appreciated!

Answer 1

why didnt you write it?

Answer 2

Because it's unpleasant algebraic terms... just hold on. I'm putting it in the equation writer.

Answer 3

So whats your problem exactly and what grade?

Answer 4

The problem:
\[\int\limits_{}^{}f^t(\sin(5^t))dt\]
\[u = 5t\]
\[du = 5^t[\log(5)]dt\]
\[\frac{1}{\log(5)}*du = 5^tdt\]

Answer 5

Blargh. ok hang on... lemme get my handy graph calculator out

Answer 6

What grade is this stuff?!

Answer 7

So...
\[\int\limits_{}^{}5^t[\sin(5^t)]dt = \int\limits_{}^{}\sin(u)*\frac{1}{\log(5)}du\]
\[= (-\cos(u))*\frac{u}{\log(5)}\]
\[= –\cos(5^t)*\frac{5^t}{\log(5)} = \frac{–5tcos(5^t)}{\log(5)}\]
That's what I got :/ But my book says it's wrong, and the only difference is that they don't have the coefficient of "5^t" in the numerator at the end, yet wolframalpha says that the antiderivative of 1/log(x) = x/log(x)... is "u" in this case not x, and wouldn't it be on the top at the end?

Answer 8

This is college Calc 1:)

Answer 9

Im still in high school geometry but ill try to help

Answer 10

|dw

Answer 11

@freckles @ganeshie8
Do you guys have a moment? :/

Answer 12

are you trying to determine if its convergent? because it is not..

Answer 13

No, I'm just evaluating the Integral:)

Answer 14

Oh duh. Sorry I am a skimmer when I read XD I like to get straight to the work... ummm um um... ok hang on. let me go back and rework it then

Answer 15

I am sorry hun, I dont understand it much myself

Answer 16

Wish I could help

Answer 17

No worries:) Anyone else?

Answer 18

Integration of sin(u)/log5 du
=-cos(u)/log5 +c

Answer 19

You multiply it by u in numerator that was the mistake

Answer 20

But why don't I consider the "unseen u^0" that's multiplied by 1 in the numerator, when integrating?

Answer 21

Is this just a formula/rule thing to memorize?

Answer 22

That' have no effect on the solution

Answer 23

u^0=1. which don't effect anything

Answer 24

Is the following not true, unless there's an identity or rule about the integration of logs?
\[\int\limits_{}^{}\sin(u)*\frac{1}{\log(5)} = \int\limits_{}^{}\sin(u)*\frac{u^0}{\log(5)} = [(-\cos(u))*\frac{u^1}{\log(5)}]\]

Answer 25

Exactly, u^0 = 1, meaning that when integrated, [u^(0+1)]/(0+1) / u^1 = u... ? Am I wrong?

Answer 26

This is done when you have only
Integration of u^0du or simply du

Answer 27

In this question it's meaningless

Answer 28

So it's a rule having to do with logs, and the antiderivatives of them?

Answer 29

Any problem now

Answer 30

I think you are clear now

Answer 31

I think... thank you for your help!

Answer 32

You are welcome

Answer 33

MY 4th comment up is the summation of my question:)

Answer 34

\[\int\limits 5^t \sin(5^t) dt \\ \text{ Let } u=5^{t} \\ \text{ then } \frac{du}{dt}=5^t \ln(5) \\ du=5^t \ln(5) dt \\ \frac{1}{\ln(5)} du=5^t dt \\ \int\limits \sin(5^t) \cdot 5^t dt =\int\limits \sin(u) \frac{1}{\ln(5)} du \\ \text{ recall } \frac{1}{\ln(5)} \text{ is a constant } \\ \text{ and this constant is a constant multiple of that one function } \sin(u) \\ \\ \text{ since } \frac{1}{\ln(5)} \text{ is a constant multiple we can bring it outside } \\ \frac{1}{\ln(5)} \int\limits \sin(u) du\]

Answer 35

also...
pretend we have
\[\int\limits u du \text{ we say this is equal \to } \frac{u^2}{2}+C \\ \text{ \not } \int\limits u \cdot 1 du=\frac{u^2}{2} \cdot u+C \text{ however} \\ \text{ to integrate a product you could apply integration by parts} \\ \text{ while that is totally not needed for this example } \\ \text{ I'm just showing a possible way to integrate a product } \\ \int\limits u \cdot 1 du=\frac{u^2}{2} \cdot 1 -\int\limits \frac{u^2}{2} \cdot u +C \\ \int\limits u \cdot 1 du=\frac{u^2}{2}-\int\limits \frac{u^3}{2}+C\]
so while you could put a one constant muliple in here it is really not helpful
in this case

I can show you a case where it is helpful
for example
\[\int\limits \ln(x) dx \\ \int\limits 1 \cdot \ln(x) dx \\ x \ln(x)-\int\limits x \frac{1}{x} dx \\ x \ln(x)-\int\limits 1 dx \\ x \ln(x)-x +C\]

but anyways in general what I'm trying to get across from my examples is:
\[\int\limits f(x) dx =\int\limits 1 \cdot f(x) dx \neq \int\limits 1 dx \cdot \int\limits f(x) dx\]

Answer 36

which is what you did

Answer 37

the slash through the equal sign means not equal if you haven't seen that before

Answer 38

oops one sec
there is no mistake above
but I meant to do that one integral the other way...
\[\int\limits u \cdot 1 du=u \cdot u -\int\limits 1 \cdot u du \\ \int\limits u \cdot 1 du=u^2-\int\limits u du\]
so actually it was helpful in that case
I just chose the wrong one to integrate

Answer 39

\[\int\limits u du = u^2 - \int\limits u du +C \\ 2 \int\limits u du=u^2+C \\ \int\limits u du =\frac{1}{2} u^2 +\frac{1}{2} C \\ \int\limits u du =\frac{1}{2} u^2+K\]

Answer 40

I don't know i you have heard of integration by parts or not

Answer 41

\[\int\limits f(x) \cdot g'(x) dx=f(x) \cdot g(x) -\int\limits f'(x) \cdot g(x) dx\]

Answer 42

Ah... this makes sense. Yes, I've heard of integration by parts, but haven't had a lot of practice yet. Thank you so much @freckles <3 ;D You rock!