Question

I am still practicing integration by parts.
If I have mistakes please catch me on it...

Answer 1

the problem is:\[\int\limits_{ }^{ }\sin^{-1}x~dx\]

Answer 2

I am going to make the second function g(x)=1, and f(x)=sin^(-1)(x).
(this is how I learned this)

So I will need to find the derivative of sin^(-1)(x).

Answer 3

\[y=\sin^{-1}(x)\]\[\sin(y)=x\]differentiating.\[y'~\cos(y)=1\]\[y'=\sec(y)\]SO,\[y'=\frac{1}{\sqrt{1-x^2}}\]

Answer 4

no I can proceed with integration.
\[\int\limits_{ }^{ } \sin^{-1}(x)~dx=x \sin^{-1}(x)- \int\limits_{ }^{ } \frac{x}{\sqrt{1-x^2}}~dx\]

Answer 5

then I am going to do a u substitution.
\[u=1-x^2\]

Answer 6

\[-\frac{1}{2}du=x~dx\]

Answer 7

\[\int\limits_{ }^{ } \sin^{-1}(x)~dx=x \sin^{-1}(x)+ \frac{1}{2}\int\limits_{ }^{ } \frac{1}{\sqrt{u}}~du\]

Answer 8

then applying power rule:

u^(-1/2) ---> 2sqrt{u}

and then we sub back the x.

Answer 9

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

Answer 10

what do you mean?

Answer 11

this is calculus. I am practicing integration by parts.

Answer 12

can you check it please?

Answer 13

you didn't learn this yet?

Answer 14

oh, I see. I'll wait for someone to come by and verify if I am right or wrong.

Answer 15

@eliassaab can you check my problem?

Answer 16

it's correct

Answer 17

ty!