<s>[INST] 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?
 [/INST] $\bf{My\; Solution::}$ Let $\displaystyle \int \sin x\cdot \sin 2x \cdot \sin 3x dx$
Using the formula 
$\bullet 2\sin A \cdot \sin B = \cos (A-B)-\cos (A+B)$
$\bullet 2\cos A \cdot \sin B = \sin (A+B)-\sin (A-B)$
So $$\displaystyle I = \frac{1}{2}\int \left[2\sin 3x \cdot \sin x\right]\cdot \sin xdx = \frac{1}{2}\int \left[\cos 2x-\cos 4x\right]\sin 2xdx$$
So $$\displaystyle I = \frac{1}{4}\int \left[2\cos 2x \cdot \sin 2x-2\cos 4x\cdot \sin 2x\right]$$
$$\displaystyle I  = \frac{1}{4}\int \left[\sin 4x-0\right]dx-\frac{1}{4}\int \left[\sin 6x-\sin 2x\right]dx$$
So $$\displaystyle I = \frac{1}{4}\left[-\frac{\cos 4x}{4}+\frac{\cos 6x}{6}-\frac{\cos 2x}{2}\right]+\mathcal{C}$$
Now $$\displaystyle \cos 6x = 1-2\sin^2(3x) = 1-2\left[3\sin x-4\sin^3 x\right]$$
and $$\displaystyle \cos 4x = 1-2\sin^2(2x) = 1-2\left[2\sin x\cdot \cos x\right]^2 = 1-4\sin^2\cdot (1-\sin^2 x)$$
and $$\displaystyle \cos 2x = 1-2\sin^2 x$$
Now put into final solution of Integral
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