Question

Anyone got clue on how to start ?? ty...question is double integral with polar coordinates....question attached on second post...ty

Answer 1

For first integral, use the substitution

u = 1+r^2
du = 2r dr

Answer 2

ok done the first integral adi.....how about the numerator?

Answer 3

also i think the hint is wrong,
the derivative of tan x is sec^2 x

Answer 4

how weird ...

Answer 5

so use the substistution hint to make cos(2x) cos^2,
then simplify the integral so you are left with sec^2
integral of sec^2 = tan x

Answer 6

We have
\[\int_{0}^{\pi/3} \int_{0}^{\sqrt{\cos2\theta}} \frac{r}{(1+r^2)^2} dr d\theta\]
Let's evaluate the integral first
Let's take
r^2= t
2rdr= dt
rdr= dt/2
when
r= \(\sqrt {\cos2\theta}\)
t= \(\cos 2\theta\)
r=0
t=0
so we get now
\[\int_{0}^{\pi/3} \int_{0}^{\cos2\theta} \frac{1}{2(1+t)^2} dt d\theta\]
we get now
\[\int_{0}^{\pi/3} [\frac{-1}{2(1+t)}]_{0}^{\cos\theta} d\theta\]
we get
\[\int_{0}^{\pi/3} -(\frac{1}{2(1+cos \theta)}-\frac{1}{2}) d\theta\]
now \(1+cos \theta=2 cos^2 \theta/2\)
so we get
\[\int_{0}^{\pi/3} -(\frac{1}{2\times \cos^2 \theta/2}-\frac{1}{2}) d\theta\]
we get
\[\int_{0}^{\pi/3} -(\frac{1}{4} \sec^2 \theta/2 -\frac{1}{2}) d\theta\]
can you do now?

Answer 7

@angalc557 Did you understand ?

Answer 8

i am reading each line you write and try to understand how you get them....wow ....the r for the numerator how it became 1/2 ? the so we get now part ty..converting to t from r that i understand...

Answer 9

oh is it from the rdr=dt/2 haha kk right?

Answer 10

Yeah , I'm here. If you don't understand anything tell me

Answer 11

The integral limits of cos2 theta then the second part it became cos theta because of?

Answer 12

Sorry I made mistake , It should be cos 2theta only. Let me rewrite

Answer 13

\[\int_{0}^{\pi /3}- (\frac{1}{2(1+\cos 2\theta)} -\frac{1}{2}) d\theta\]
\[1+\cos 2\theta= 2 \cos^2 \theta\]
so
\[\int_{0}^{\pi /3}- (\frac{1}{2(2 \cos^2 \theta)} -\frac{1}{2}) d\theta\]
we get

\[\int_{0}^{\pi /3}- (\frac{\sec^2 \theta}{4} -\frac{1}{2}) d\theta\]
Can you do it now?

Answer 14

@angalc557 did you understand?

Answer 15

i see....kk still going through.....so far so good...haha

Answer 16

@ash2326 is the final answer pi/6-sqrt3/4 ty...??

Answer 17

@angalc557 let me check,

Answer 18

yeah correct
great work @angalc557

Answer 19

haha ty quite challenging without help...:)