Question

Integration..q below

Answer 1

\[\int\limits_{0}^{1/2} \frac{ \ln(1+2x) dx}{ 1+4x ^{2} }\]

Answer 2

i took\[x=\cos ^{2}\]

Answer 3

*\[\cos ^{2}\theta \]

Answer 4

so i got..
\[\int\limits_{\pi/4}^{\pi/2}\frac{ \ln(\cos2\theta)\sin2\theta d \theta }{1+4\cos ^{4}\theta }\]

Answer 5

how to simplify further?

Answer 6

@hartnn @iambatman @dan815 @freckles @UnkleRhaukus @ganeshie8

Answer 7

maybe you can use
\(\int \limits_a^b f(x)dx = \int \limits_a^b f(a+b-x) dx\)

i think it can simplify your integral significantly

Answer 8

but a+b here is 3pi/4...

Answer 9

oh wait..i think i have made a mistake..

Answer 10

for the original integral with limits 0 to 1/2

Answer 11

i think..its getting further complicated if i apply the rule to the original integral..

Answer 12

if i apply the rule for the trig integral..the numerator is easily simplified..but not the denominator..as it doesn't have "2(theta)"..

Answer 13

@hartnn ?

Answer 14

thinking of another substitution...
maybe we can bring the integral in the form of
\(\int \limits_0^\pi \ln \sin x\)
which is a classic one

Answer 15

hmm..

Answer 16

if 1+2x = sin y
what will be 1+4x^2 ?
and dx?

Answer 17

sin^2 y+1-2siny

Answer 18

and cosy/2 resp

Answer 19

yeah, that didn't turn out as easy as I initially thought

Answer 20

u mean u integrated with this substitution..?

Answer 21

@hartnn ?

Answer 22

Please try this variable change:
\[\Large 1 + 2x = {e^z}\]

Answer 23

where \(z\) is the new variable of integration

Answer 24

i tried it..this is what i got..
\[1/2\int\limits_{0}^{1/2}\frac{ z(e ^{z})dz }{ (e ^{z}-1)^2 }\]

Answer 25

@Michele_Laino ?

Answer 26

I'm very sorry I have made a big error!!

Answer 27

oh..what error?

Answer 28

an error of substitution

Answer 29

oh..its ok.. i am trying diff. substitutions..but i am not able to find an easy one..

Answer 30

@IrishBoy123 please help!

Answer 31

do u have any idea to further simply my trig integral..?

Answer 32

I have tried the integration by part, nevertheless I haven't obtained any result

Answer 33

hmm..

Answer 34

michele

it's definite

have you thought about one of those Feynman tricks???!!!

Answer 35

no, sincerely I have not thought about them

Answer 36

Hii

Answer 37

if we change variable: \(z=2x\), and we apply the integration by parts, we get:
\[\large \int {\frac{{\ln \left( {1 + 2x} \right)}}{{1 + 4{x^2}}}dx = \frac{1}{2}\left( {\arctan z} \right)\ln \left( {1 + z} \right)} - \frac{1}{2}\int {\arctan z\frac{{dz}}{{1 + z}}} \]

Answer 38

i got a clue from ur method!

Answer 39

2x=tan(theta)!!

Answer 40

one sec..seeing if it works..

Answer 41

Nice :) for sure that works !
just for a variety, you may try this :

\[\int\limits_{0}^{1/2} \frac{ \ln(1+2x) dx}{ 1+4x ^{2} } = \frac{1}{2}\int\limits_{0}^{1/2} \int\limits_{y}^{1/2} \frac{ 1}{ (1+2y)(1+4x^{2}) } \,dydx\]

Answer 42

@rsadhvika ..i have not been taught abt double integral..

Answer 43

that is okay, its not hard and i guess you don't really need to be taught to understand it :)
just treat it as an iterated integral : integrate the inside with respect to y first, then whatever you get, you integrate that with respect to x

Answer 44

oh its that simple..!thanks

Answer 45

thank u @Michele_Laino !

Answer 46

honestly i don't want this thread to be ur introduction to double integrals, maybe just go ahread with IBP...

Answer 47

:)

Answer 48

thanks all..!

Answer 49

i have another q..can anyone help?

Answer 50

i will post it separately..

Answer 51

okay sure :) where do you find such nice integral questions