Question

integrate

Answer 1

? QUE

Answer 2

\[\huge \int\limits \frac{xsin^{-1}x}{\sqrt{1-x^2} }\]

Answer 3

dx

Answer 4

@lgbasallote

Answer 5

Use this substitution,\[\arcsin x =t\]

Answer 6

let u = sin^-1 x
du = 1/sqr(1 - x^2) dx

x = sin u so...

\(\int (sin u)u du\)

Answer 7

integration by part

s = u dw = sinudu
ds = du w = -cosu

-ucosu + \(\int cosudu\)
-ucosu + sinu

sub back

Answer 8

u = sin^-1 x
cos u = sqrt (1 - x^2)
sin u = x

so....

\((\sin^{-1} x)\)( \(\sqrt{1-x^2}\)) + x

Answer 9

thx a lot @lgbasallote

Answer 10

thx evryone

Answer 11

shouldn't it be \[-\sin^{-1}x \sqrt{1-x^2}+x+C ?, i mean with the negative sign\] I mean, with negative sign

Answer 12

another solution:

Integration by parts:
\[u=\sin^{-1} x\]\[du=1/(\sqrt{1-x^2})dx\]\[\int\limits_{}^{}dv = \int\limits_{}^{}[x/\sqrt{(1-x^2})]dx\]\[v= -\sqrt{1-x^2}\]

Note: \[\int\limits_{}^{}udv= uv - \int\limits_{}^{}vdu\]

Therefore: \[\int\limits_{}^{}(x \sin^{-1} x)/(\sqrt{1-x^2}) dx = -\sin^{-1} x \sqrt{1-x^2} + x +C\]

Answer 13

yup with negative and constant..thanks for pointing that out