Why I am getting different answer? I have just started learning single variable calculus. I'm confused in a problem from sometime. I didn't get why my answer is different from the book.
$$
\require{cancel}
\begin{align}
&\int\sin x \sin 2x \sin 3x\,dx\\
&=\int\sin x\;\,2\sin x\cos x \left(3\sin x - 4\sin^3 x\right)\,dx\\
&\qquad\text{Let }\sin x = t, \text{ then}\\
&\qquad\quad\cos x\, dx = dt\\
&=\int t\;2t\left(3t - 4 t^3\right)\,dt\\
&=\int 2t^2\left(3t - 4t^3\right)\,dt\\
&=\int\left(6t^3-8t^5\right)\,dt\\
&=6\int t^3\,dt - 8\int t^5\,dt\\
&=\cancel{6}\,3\frac{t^4}{\cancel{4}2}+c_1-\cancel{8}\,4\frac{t^6}{\cancel{6}3}+c_2\\
&=\frac32t^4-\frac43t^6+C\\
&=\frac32\sin^4x-\frac43\sin^6x+C
\end{align}
$$
The answer given in my book is 
$$\displaystyle\frac{1}{4}\left[\frac{1}{6}\cos 6x - \frac{1}{4}\cos 4x - \frac{1}{2}\cos 2x\right]  + C. \ $$
Where did I go wrong?

Both solutions are correct, they only differ by a constant. See also Wolfram Alpha.
This is similar to the following situation: Both $f(x) = \sin^2(x)$ and $g(x) = -\cos^2(x)$ are antiderivatives of $2\sin(x)\cos(x)$. Even if they look quite different, they only differ by a constant: $f(x) = 1 + g(x)$.
