Question

small calculus .... /..

Answer 1

\[\int\limits ( \sin^{-1} (\frac{ 2tanx }{1+\tan ^{2} x }))dx\]

the bracket after simplifying got (sin 2x )

Answer 2

\[\int\limits \sin^{-1} (\sin2x)dx\] .. ??

Answer 3

its sin 2x ....how is it possible

Answer 4

@reemii

Answer 5

well not clear ...

Answer 6

is \(\sin^{-1}x = \frac1{\sin x}\) or \(\arcsin x\) ?

Answer 7

arc sinx

Answer 8

well both r correct ....rt

Answer 9

I would start by integration by parts. Make \(x\) appea with the "usual trick" : \(\int f(x)dx = xf(x) - \int xf'(x) dx\). Did you try that?

Answer 10

noo i dont like by parts ..... it takes lot of time i did not try that ,,,,,,,,

Answer 11

the derivative of \(\arcsin u = \frac1{\sqrt{1-u^2}}u'\). This might give something interesting (\(u=\sin(2x))\).

Answer 12

well we r gona take derivative of sin2x

Answer 13

@mathslover @mathstudent55 @mind2000 @amistre64

Answer 14

what must sin-1(2inx) be?

Answer 15

you forgot the power