Question

question attached below

Answer 1

solve \[\int\limits_{0}^{\pi} 1/(2x+\pi) dx\]

Answer 2

substitute u=2x+pi; du=2dx

Answer 3

@deeprony7 can you please show me the steps?

Answer 4

i think you do this integral more or less in your head yes?

Answer 5

@ satellite yes but i mean the steps to the solution because i got a different answer to the solution answer given

Answer 6

you have
\[\int\frac{1}{u} du\] get \[\ln(u)\] and adjusting for the 2 you get
\[\frac{1}{2}\ln(2x+\pi)\] unless i made a bush league mistake

Answer 7

sorry. then deeprony had the u - sub

Answer 8

(1/2)(ln|2∗π+π|−ln|2∗0+π|)=(1/2)(ln|3π|−ln|π|)=1/2ln(3)

Answer 9

no worries @ satellite

Answer 10

put
\[u=2x+\pi, dx=\frac{1}{2}du\]
\[u(0)=\pi, u(\pi)=3\pi\]

Answer 11

THANKYOU @ tatiana but i'm confused at how you got the denominater of 2ln3?

Answer 12

\[\frac{1}{2}\int _{\pi}^{3\pi} \frac{du}{u}=\frac{1}{2}\ln(u)|_{-pi}^{3\pi}\]

Answer 13

\[=\frac{1}{2}(\ln(3\pi)-\ln(\pi))\]
\[=\frac{1}{2}\ln(\frac{3\pi}{\pi})\]
\[=\frac{1}{2}\ln(3)\]

Answer 14

OHHHHHHHHHHHHHHHHHHHHHHHHHHHHH I GET where 2ln(3) comes from!!!!!!!!!!!

Answer 15

YAYAYAYAYAY! THANKS SOSOSO MUCH!