Question

Integrate!

Answer 1

\[\LARGE \int\limits\limits_0^\pi \sin^{-1}\sqrt \frac{x}{a+x}dx\]

Answer 2

Attempt:
Let x=a tan^2 theta..

Answer 3

correct hai

Answer 4

but i am confused about limits?
I am getting lower limit as 0 and upper as
\[\LARGE \tan^{-1} \sqrt{\frac{\pi}{a}}\]
lol

Answer 5

tan inverse function ka koi standard formula nai hai ?

Answer 6

dunno..

Answer 7

I am getting
\[\Huge \theta]_0^{\tan^{-1}\sqrt{\frac{\pi}{a}}}\]

Answer 8

no actually theta^2/2

Answer 9

\[\LARGE \frac{1}{2}\tan^{-1} \frac{\pi}{a}\]

Answer 10

uska square
weird
what should be the answer ?

Answer 11

\[\LARGE \frac{a(\pi-2)}{2}\]

Answer 12

wait i didnt change dx :P

Answer 13

i suspect uv rule :O

Answer 14

one way or the other, you'll need product rule
\(


\int \arctan x dx =\arctan x - \frac{1}{2}\ln\left(1+x^2\right) + C\\
\)

Answer 15

see i am getting this
\[\LARGE 2a \int\limits \theta \sec^2 \theta \tan \theta d \theta\]
how do i do with this?

Answer 16

let tan theta=t?

Answer 17

2a integration t theta d theta

Answer 18

then i can write theta in terms of t
\[2a \int\limits t . \tan^{-1}t dt\]
how ?by parts now?HOW...

Answer 19

2a sec^2 theta tan theta

Answer 20

then tan theta=t

Answer 21

the very original function would be tan inv sqrt (x/a)
wouldn't it be better to do parts here ?

Answer 22

how?

Answer 23

which step?

Answer 24

you go theta na for sin inv ....
and theta = ....

Answer 25

but dx bhi to kia

Answer 26

int tan inv sqrt (x/a) dx
haven't tried....
see which one is easy for you to do parts

Answer 27

hey if i let x=a then dx=0? :P

Answer 28

then integration=0

Answer 29

you cannot let variable = constant