Question

Can somebody help please !!!!!!!!!!!!! I am completely stuck, don't even know where to start !!
Find each of the following indefinite integrals, identifying any general rules of calculus that you use.
integral sign [sec^2*(4lnx)] \ x (e^-pi\8 less than x less than e^pi\8 )
Thank you millions for anyone who could help !!

Answer 1

Have you tried typing formulas, equations, symbols, etc., into Equation Editor? If so, has it worked for you? If not, perhaps you could DRAW your integral to make it easier for a potential helper to decipher.

Answer 2

so your question is

\(\Large \int \frac{\sec^2(4\ln x)}{x}\) for \(\Large e^{\frac{-\pi}{8}}\le x \le e^{\frac{\pi}{8}} \)

Answer 3

is this your question

Answer 4

I have tried but it didn't work ! sorry I am new to this site and only just learning how to use it, but I will try again !
Yes that is my question, but is less than x not equal on both side of x
Thanks very much !!

Answer 5

\[\int\limits sign [\sec^2*(4lnx)] \ x \rightarrow \int\limits_{}^{}(\sec ^{2}x)(4 \ln x) dx\] ???? Jonask was kind enough to try to verify what you meant.

Would you please respond.

Answer 6

thanks mathmale

but for now we are not sure what is your question
\(\Large \int \dfrac{\sec^2(4\ln x)}{x}\) for \(\Large e^{\frac{-\pi}{8}}<x < e^{\frac{\pi}{8}} \)

OR
\(\Large \int \sec^2(x)(4\ln x)\) for \(\Large e^{\frac{-\pi}{8}}< x < e^{\frac{\pi}{8}} \)

Answer 7

The top one please !! thank you !!

Answer 8

Again I have to ask for verification: What is the argument of your sec function? Is the argument x, or is the argument 4 ln x?

See why it's important to post your question the clearest form possible?

Answer 9

let \(t=\ln x\) the we see that \(dt=\dfrac{dx}{x}=\dfrac{dx}{e^t}\implies e^tdt=dx\)
you need to substitute this in the integral,since x is positive ie

\(\Large \int \frac{\sec^2(4\ln x)}{x}dx=\int\sec^2(4t)dt=\frac{1}{4}\tan (4t)\)

Answer 10

\(\frac{1}{4}\tan( 4\ln x)\)

Answer 11

I am very sorry if I confused you, as I said I have tried to use the signs but they don't seem to work for me or I do something wrong. Jonask has replied to me asking which one of two was my question, and I have replied that is the top one. Thank you for helping !

Answer 12

for future purposes ,u shud post like this ,this is how i posted the latex

\ [ \int \frac{\sec^2(4\ln x)}{x}\ ] for
\ [ \large e^{\frac{-\pi}{8}}\le x \le e^{\frac{\pi}{8}} \]

which gives

\[ \int \frac{\sec^2(4\ln x)}{x}\] for
\[ \large e^{\frac{-\pi}{8}}\le x \le e^{\frac{\pi}{8}} \]

\ [ shud be attached like this \[

Answer 13

I'm going to assume that Jonask's interpretation of the integral is the correct one. (See the illustration, above.) I'll move that " 4 " outside of the integral to simplify the integration.

Next, I look for a possible substitution, and see that if I were to choose u=ln x to be that substitution, then du would be du = dx/x. Can you agree with this?

Making that substitution results in:\[4\int\limits_{}^{}\sec ^{2}u*du\]

Answer 14

Can you, guzu, integrate this?

Answer 15

Wao , thanks very much Jonask, I will try !

Answer 16

\[int\frac{\sec^2(4\lnx)}{x}\]
thank you jonask for showing this to me, but how was I supposed to know know how to type this in???

Answer 17

Jonask, would you mind awfully to explain please, where has the X gone from the denominator in the first step you made. Thanks very much !!

Answer 18

okay
we start by using this substitution
\(t=\ln x\) when you take the derive you get \(dt=\color{blue}{\dfrac{dx}{x}}\)
i will remove the blue part with \(dt\) ,this is why we call it substitution
\[ \int \frac{\sec^2(4\ln x)}{x}dx=\int\sec^2(4\ln x)\color{blue}{\dfrac{dx}{x}}=\int \sec^2(4t)dt=\frac{1}{4} \tan(4 \ln x )\]

Answer 19

Thank you !!

Answer 20

yw ,did u understand?

Answer 21

yes, I do now ! thank you !

Answer 22

your welcome