Question

Evaluate integral of( x sin^{-1}x)/sqrt(1-x^{2})dx

Answer 1

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

Answer 2

\[\int\limits_{}^{} \frac{xsin^-{1}x}{\sqrt{1-x^2}}dx\]

Answer 3

\[\int\limits_{}^{}\frac{x}{\sqrt{1-x^2}} \cdot \sin^{-1}(x) dx \text{ use integration by parts }\]

----------------------------
\[\int\limits_{}^{}\frac{x}{\sqrt{1-x^2}} dx \text{ } u=1-x^2 => du=-2 x dx => \frac{-1}{2} du=x dx\]
\[\int\limits_{}^{}\frac{\frac{-1}{2} du}{\sqrt{u}}=\frac{-1}{2}\int\limits_{}^{} u^\frac{-1}{2} du\]
I will let you finish this part
-----------------------------
Then ask yourself what is (arcsin(x))'

See how this works out for yea

Let me know if you need further assistance

Answer 4

Integrate by parts.
\[
u=\sin^{-1} x ,\, dv= \frac x{\sqrt{1-x^2}}

\]
Work it out

Answer 5

\[du =\frac 1 {\sqrt{1-x^2}}\\
v= -\sqrt{1-x^2}

\]
Compute now
\[
u v - \int v du

\]
and you will be done.

Answer 6

\[x- \sqrt{1-x^2}\sin^-1x\] +c

Answer 7

thank you sooo much..

Answer 8

but how did you get\[v=-\sqrt{1-x^2}\]

Answer 9

did you see my post above?

Answer 10

if you follow that you will see

Answer 11

yes

Answer 12

hld on pls i'm wrkng on it