Question

integral (pi/4 to pi/2) of cos(x)/sin(x)

Answer 1

\[\int\limits_{\pi/4}^{\pi/2}\frac{ \cos(x) }{ \sin(x) }\]

Answer 2

since cos is a derivative of sine ....

Answer 3

and i know you use substitution and you insert u=sin(x) and d(u)=cos(x) and it turns into \[\int\limits_{?}^{?}\frac{ 1 }{ u }\] and then it turns into \[\ln \left| u \right|\] and then it turns into \[\ln \sin(x) from \frac{ \pi }{ 4 } \to \frac{ \pi }{ 2}\]

Answer 4

but you plug it in and you get \[0-\ln \frac{ \sqrt{2} }{ 2 }\]

Answer 5

but i think this is wrong

Answer 6

its correct, it just needs simplification.

Answer 7

how?

Answer 8

what is \(\large \dfrac{2}{\sqrt2}=...?\)
can you simplify this ?
and write it in the form 2^(.....)

Answer 9

You can also use the property of logarithm that \[\ln{\frac{a}{b}}=\ln a -\ln b\]

Answer 10

and \[\ln a^b = b\ln a\] as well

Answer 11

\[\frac{ 1 }{ 2 } \ln 2 - \ln 2\]

Answer 12

the log property that you'll need is \(\large \log a^b=b\log a\)

Answer 13

They are fundamentally the same.

Answer 14

if we keep it in u

u = sinx ; x=pi/4 .... u = sqrt(2)/2
u = sinx; x=pi/2 .... u = 1

1 - ln(sqrt(2)/2) is correct

Answer 15

i dont really understand

Answer 16

did you forget - *minus*

Answer 17

i dont think it needs to be "simplified" any. the calculator knows how to work it as is :)

Answer 18

ln 1 =0

Answer 19

.... i meant that lol

Answer 20

its not in the multiple choice

Answer 21

\[-\ln (\frac{\sqrt 2}{2})=-\ln(\frac{1}{\sqrt{2}})=-\ln(2)^{-\frac{1}{2}}=\frac{\ln 2}{2}\]

Answer 22

whats in the options then?

Answer 23

we can manipulate this thing all day, but it might go quicker if we knew the options

Answer 24

that was negative sign before also
\(-(\frac{ 1 }{ 2 } \ln 2 - \ln 2)= \ln 2- \frac{ 1 }{ 2 } \ln 2 = \frac{ 1 }{ 2 } \ln 2\)

Answer 25

A

Answer 26

how

Answer 27

\(\frac{ 1 }{ 2 } \ln 2 = \ln 2^{\frac{ 1 }{ 2 } }=\ln \sqrt2\)

Answer 28

i used the same log property
\(\large \log a^b=b\log a\)

Answer 29

did you get it ?

Answer 30

yup thanks

Answer 31

welcome ^_^