Integration problem no. 2

[mathwannabe]

Hello everybody

Here it is...
I managed to do this, but my solution does not match the one proided by the book possibly because my trig sucks big time and I havent been able to tidy up the final expression. Anyway, here is my work:

\(\displaystyle \displaystyle\int\dfrac{2x-\sqrt{\arcsin{x}}}{\sqrt{1-x^2}}dx=?\)

\(\displaystyle x=\sin{t}\) => \(\displaystyle t=\arcsin{x}\)
\(\displaystyle dx=\cos{t}\;dt\)

\(\displaystyle \displaystyle\int\dfrac{2\sin{t}-\sqrt{\arcsin{(\sin{t})}}}{\sqrt{1-\sin^2{t}}}\cos{t}\;dt=\)

\(\displaystyle \displaystyle\int\dfrac{2\sin{t}-\sqrt{t}}{\sqrt{\cos^2{t}}}\cos{t}\;dt=\)

\(\displaystyle \displaystyle\int\dfrac{2\sin{t}-t^{1/2}}{\cos{t}}\cos{t}\;dt=\)

\(\displaystyle \displaystyle\int2\sin{t}-t^{1/2}\;dt=\)

\(\displaystyle \displaystyle\int2\sin{t}\;dt-\int t^{1/2}\;dt=\)

\(\displaystyle 2\cos{t}-\frac{2}{3}t^{3/2}+C=\)

\(\displaystyle 2\cos{(\arcsin{x})}-\frac{2}{3}\sqrt{(\arcsin{x})^3}+C\) <========== This one is my answer.

\(\displaystyle C-2\sqrt{1-x^2}-\frac{2}{3}\sqrt{(\arcsin{x})^3}\) <========== This one is from the book. The book offers only final answers, without the steps.

Did I do the integral wrong, or have I just failed to transform the final expression? If yes, how can I transform it?

 

[Deleted member 4993]

Hello everybody

Here it is...
I managed to do this, but my solution does not match the one proided by the book possibly because my trig sucks big time and I havent been able to tidy up the final expression. Anyway, here is my work:

\(\displaystyle \int\dfrac{2x-\sqrt{\arcsin{x}}}{\sqrt{1-x^2}}dx=?\)

\(\displaystyle x=\sin{t}\) => \(\displaystyle t=\arcsin{x}\)
\(\displaystyle dx=\cos{t}\;dt\)

\(\displaystyle \int\dfrac{2\sin{t}-\sqrt{\arcsin{(\sin{t})}}}{\sqrt{1-\sin^2{t}}}\cos{t}\;dt=\)

\(\displaystyle \int\dfrac{2\sin{t}-\sqrt{t}}{\sqrt{\cos^2{t}}}\cos{t}\;dt=\)

\(\displaystyle \int\dfrac{2\sin{t}-t^{1/2}}{\cos{t}}\cos{t}\;dt=\)

\(\displaystyle \int2\sin{t}-t^{1/2}\;dt=\)

\(\displaystyle \int2\sin{t}\;dt-\int t^{1/2}\;dt=\)

\(\displaystyle 2\cos{t}-\frac{2}{3}t^{3/2}+C=\)

\(\displaystyle 2\cos{(\arcsin{x})}-\frac{2}{3}\sqrt{(\arcsin{x})^3}+C\) <========== This one is my answer.

\(\displaystyle C-2\sqrt{1-x^2}-\frac{2}{3}\sqrt{(\arcsin{x})^3}\) <========== This one is from the book. The book offers only final answers, without the steps.

Did I do the integral wrong, or have I just failed to transform the final expression? If yes, how can I transform it?

To verify your answer:

differentiate \(\displaystyle 2\cos{(\arcsin{x})}\)

and

differentiate \(\displaystyle C-2\sqrt{1-x^2}\)

do those come out to be same?

If those do - you are correct!!

 

[DrPhil]

Hello everybody

Here it is...
I managed to do this, but my solution does not match the one proided by the book possibly because my trig sucks big time and I havent been able to tidy up the final expression. Anyway, here is my work:

\(\displaystyle \int\dfrac{2x-\sqrt{\arcsin{x}}}{\sqrt{1-x^2}}dx=?\)

. . .

\(\displaystyle 2\cos{(\arcsin{x})}-\frac{2}{3}\sqrt{(\arcsin{x})^3}+C\) <========== This one is my answer.

\(\displaystyle C-2\sqrt{1-x^2}-\frac{2}{3}\sqrt{(\arcsin{x})^3}\) <========== This one is from the book. The book offers only final answers, without the steps.

Did I do the integral wrong, or have I just failed to transform the final expression? If yes, how can I transform it?

The first suggestion I would have made is to split the integrand into two parts initially:

\(\displaystyle \displaystyle\int\dfrac{2x}{\sqrt{1-x^2}}dx - \int\dfrac {\sqrt{\arcsin{x}}}{\sqrt{1-x^2}}dx\)

My second comment is that you can use the unit circle to find cos(arcsin(x)) = sqrt(1 - x^2)

Good work!

 

[mathwannabe]

The first suggestion I would have made is to split the integrand into two parts initially:

\(\displaystyle \displaystyle\int\dfrac{2x}{\sqrt{1-x^2}}dx - \int\dfrac {\sqrt{\arcsin{x}}}{\sqrt{1-x^2}}dx\)

My second comment is that you can use the unit circle to find cos(arcsin(x)) = sqrt(1 - x^2)

Good work!

Thank you

 

[mathwannabe]

To verify your answer:

differentiate \(\displaystyle 2\cos{(\arcsin{x})}\)

and

differentiate \(\displaystyle C-2\sqrt{1-x^2}\)

do those come out to be same?

If those do - you are correct!!

It does! :3

 

[soroban]

Hello, mathwannabe!

\(\displaystyle \displaystyle\int\dfrac{2x-\sqrt{\arcsin{x}}}{\sqrt{1-x^2}}dx\)

Make two integrals:

. . \(\displaystyle \displaystyle \int\frac{2x}{\sqrt{1-x^2}}\,dx - \int\frac{\sqrt{\arcsin x}}{\sqrt{1-x^2}}\,dx\)

. . \(\displaystyle \displaystyle =\;\int(1-x^2)^{-\frac{1}{2}}(2x\,dx) \:-\:\int (\arcsin x)^{\frac{1}{2}}\frac{dx}{\sqrt{1-x^2}} \)


First integral: .\(\displaystyle u \:=\:1-x^2 \quad\Rightarrow\quad du \:=\:\text{-}2x\,dx \quad\Rightarrow\quad 2x\,dx \,=\,\text{-}du\)

Second integral: .\(\displaystyle v \,=\,\arcsin x \quad\Rightarrow\quad dv \,=\,\frac{dx}{\sqrt{1-x^2}}\)

Substitute and we have: .\(\displaystyle \displaystyle -\!\!\int u^{-\frac{1}{2}}\,du \:-\:\int v^{\frac{1}{2}}\,dv \)

Can you finish it?