Where I'm wrong? Why my answer is different from the book? The question is to integrate 
$$\int x\cos^{-1}x dx$$
Answer in my book
$$(2x^2-1)\frac{\cos^{-1}x}{4}-\frac{x}{4}\sqrt{1-x^2}+C$$
I'm learning single variable calculus right now and at current about integration with part. I'm confused in a problem from sometime. I don't know where I am wrong. Please have a look at the images.

Solution. $\newcommand{\dd}{\; \mathrm{d}}$
  $$
\begin{align}
&\cos^{-1} x \int x \dd x - \int \left[\frac{-1}{\sqrt{1-x^2}}\cdot\frac{x^2}2\right] \dd x=\\
=&\cos^{-1}x \cdot \frac{x^2}2 + \int \frac{x^2}{2\sqrt{1-x^2}} \dd x2=\\
=&\frac{x^2}2 \cdot \cos^{-1}x + \frac12 \int \frac{x}{\sqrt{1-x^2}} \dd x =
\begin{vmatrix}x=\sin t\\ \dd x=\cos t\dd t\end{vmatrix} = \\
=&\frac{x^2\cos^{-1}x}2 + \frac12 \int \frac{\sin^2t \cos t}{\sqrt{1-\sin^2t}} \dd t =\\
=&\frac{x^2\cos^{-1}x}2 + \frac12 \int \frac{\sin^2t \cos t}{\cos t} \dd t =\\
=&\frac{x^2\cos^{-1}x}2 + \frac12 \int \sin^2t \dd t =\\
=&\frac{x^2\cos^{-1}x}2 + \frac12 \int \frac{1-\cos2t}2 \dd t =\\
=&\frac{x^2\cos^{-1}x}2 + \frac12 \int \frac12 \dd t -\frac 14 \int \cos2t \dd t =\\
=&\frac{x^2\cos^{-1}x}2 + \frac14 t - \frac18 \sin 2t + C =\\
=&\frac{x^2\cos^{-1}x}2 + \frac14 \sin^{-1}x - \frac18 \sin2\sin^{-1}x + C =\\
=&\frac{x^2\cos^{-1}x}2 + \frac14 \sin^{-1}x - \frac18 \sin(2\sin^{-1}x) + C
\end{align}
$$ 

Please help. Thankyou in advance. 

Your answer is not false, but you didn't thoroughly simplify.
This is why both answers are the same:


*

*$\arcsin x+\arccos x=\dfrac\pi2$

*$\sin(2\arcsin x)=2\sin(\arcsin x)\cos(\arcsin x)=2x\sqrt{1-x^2}.$

