Question

Integrate!

Answer 1

\[\LARGE \int\limits_{-a}^{+a} \sqrt{\frac{a-x}{a+x}}dx\]

Answer 2

I tried putting x=acos2theta

Answer 3

I am confused that WILL we change the limits?

Answer 4

why did you put that substitution ?
what function will u get after the substitution ?

Answer 5

\[\large \int\limits \sin^2 \theta d \theta\]

Answer 6

why i am i not getting the same...
also, if you are changing dx to dtheta, then you MUST change the limits

Answer 7

limits change hongi?

Answer 8

if you found dtheta, then you must change limits too

Answer 9

but after changing limits im getting weird results lol
lower limit pi/2 upper 0
how is it possible?

Answer 10

how is that weird :O

Answer 11

is that integration? upper limit is generally>lower limits..how is this integration then
we are not practically "Integrating"

Answer 12

\[acos2\theta=-a=>\cos2 \theta=-1 =>2 \theta= \pi=>\theta=\frac{\pi}{2}\]
how pi?

Answer 13

\(\int \limits_a^b =-\int \limits_b^a \)
pi/2 is correct

Answer 14

upper is 0

Answer 15

yeah limits are correct

Answer 16

hello @hartnn

Answer 17

abeee, idhar hi to hu

Answer 18

..i dont think so

Answer 19

i did say limits are correct...

Answer 20

what the doubt ?

Answer 21

integration answer will come -ve and wrong i tried it..

Answer 22

the function sqrt (....) simplifies to tan theta ? right ?

Answer 23

\[\int\limits \tan \theta \times -2a \sin 2 \theta d \theta=>- 4a \int\limits \tan \theta \sin \theta \cos \theta= -4a \int\limits \sin^2 \theta d \theta\]

Answer 24

\[\frac{1}{2}(\theta-\sin \theta \cos \theta) ]_\frac{\pi}{2}^0\]

Answer 25

-2a int 1-2cos2theta
-2a (t - sin 2t) 0 to pi/2
2a (pi/2 - 0 )
a pi
?

Answer 26

\[−2a (\frac{\pi}{2})=-a \pi\]

Answer 27

mera to api aaya

Answer 28

0 to pi/2 kaise kara? - hatake reverse limits?

Answer 29

oops, yes, i get - api too

Answer 30

:/

Answer 31

hogaya api agaya..- lagakar change limits
(-)(-)=+

Answer 32

-2a int 1-2cos2theta
-2a (t - sin 2t) pi/2 to 0
2a (pi/2 - 0 )
a pi
this is what u did too, right ?

Answer 33

i think yes, and you are correct, and we both now get api

Answer 34

:D