TITLE: Why I am getting different answer?
QUESTION [6 upvotes]: 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?

REPLY [9 votes]: $
\renewcommand{\a}{\alpha}
\renewcommand{\b}{\beta}
$
Using trigonometric identities
$$
\begin{aligned}
\sin \a \sin \b 
& = \frac{1}{2}\Big( \cos\left(\a-\b\right)- \cos\left(\a+\b\right)\Big) \\
\sin 3\a 
& =  3 \sin x - 4\sin^3 x 
, 
\end{aligned}
$$
we write 
$$
\begin{aligned}
\int \sin x  \sin 2x   \sin 3x \,  dx 
& = \frac{1}{2} \int  \sin 3x \, \Big( \cos x - \cos 3x  \Big)\,  dx 
\\
& = \frac{1}{2}\int \sin 3x \cos x \, dx - \frac{1}{2} \int\sin 3x \cos 3x \,d(x) 
\\
& = \frac{1}{2}\int \big( 3 \sin x - 4\sin^3 x \big) \cos x \, dx - \frac{1}{6} \int\sin 3x \cos 3x \,d(3x)
\\
& = \frac{1}{2}\int \big( 3 \sin x - 4\sin^3 x \big) \, d(\sin x) - \frac{1}{6} \int\sin 3x   \,d(\sin 3x)
\\
& = \frac{3}{4}\sin^2 x - \frac{1}{2}\sin^4 x  - \frac{1}{12} \sin^2 3x + C
\\
& =  \frac{3}{4}\sin^2 x - \frac{1}{2}\sin^4 x  - \frac{1}{12} \big( 3 \sin x - 4\sin^3 x \big)^2 + C
\\ 
& =  \frac{3}{4}\sin^2 x - \frac{1}{2}\sin^4 x  - \frac{1}{12} \big( 9 \sin^2 x - 24 \sin ^4 x + 16\sin^6 x \big)  + C
\\
& =  \frac{3}{2}\sin^4 x  - \frac{4}{3}  \sin^6 x   + C 
\end{aligned}
$$

Let us now make sure that our answer matches the textbook one $$\displaystyle\frac{1}{4}\left[\frac{1}{6}\cos 6x - \frac{1}{4}\cos 4x - \frac{1}{2}\cos 2x\right]  + C = \frac{1}{24}\cos 6x - \frac{1}{16}\cos 4x - \frac{1}{8}\cos 2x  + C$$
Indeed, observe that 
$$
\begin{aligned}
\cos 6x &= \cos \big(2(3x) \big) 
\\ & = \cos^2 3x - \sin^2 3x 
\\ & = 1 - 2 \sin^2 3x 
\\ & = 1 - 2 \big(3\sin x - 4\sin^3 x\big)^2
\\ & = 1 - 2 \big(9\sin^2 x - 24\sin^4 x + 16\sin^6 x\big)
\\ & = 1 - 18\sin^2 x + 48 \sin^4 x - 32\sin^6 x 
\\
\cos 4x &= \cos \big(2(2x) \big) 
\\ & = \cos^2 2x - \sin^2 2x 
\\ & = 1 - 2 \sin^2 2x 
\\ & = 1 - 2 \big(2\sin x \cos x \big)^2
\\ & = 1 - 2 \big(4 \sin^2 x \cos^2 x\big)
\\ & = 1 - 8\sin^2 x \left(1  - \sin^2 x \right) 
\\ & = 1 - 8\sin^2 x +  8 \sin^4 x
\\
\cos 2x &= \cos^2 x  - \sin^2 x 
\\ & = 1 - 2 \sin ^2 x
\end{aligned}
$$
Substituting everything back into the textbook answer, we get
$$
\begin{aligned}
\frac{1}{24}\cos 6x - \frac{1}{16}\cos 4x - \frac{1}{8}\cos 2x  + C 
= \ & 
\frac{1}{24} \left(1 - 18\sin^2 x + 48 \sin^4 x - 32\sin^6 x \right) - 
\\ - \ &
\frac{1}{16} \left( 1 - 8\sin^2 x +  8 \sin^4 x\right) -
\frac{1}{8} \left(1 - 2\sin^2 x  \right) +
C =
\\ = \  &
\left(\frac{1}{24} - \frac{1}{16} - \frac{1}{8} \right) -
\left(\frac{18}{24} - \frac{8}{16} - \frac{2}{8} \right) \sin^2 x +
\\ +\ &
\left(\frac{48}{24} - \frac{4}{16} \right) \sin^4 x -
\frac{32}{24} \sin^6 x +
C  =
\\ = \ &
\left(C - \frac{7}{48}\right) -
\left(\frac{3}{4} - \frac{1}{2} - \frac{1}{4} \right) \sin^2 x +
\left(2 - \frac{1}{2} \right) \sin^4 x -
\frac{4}{3} \sin^6 x 
\\ = \ &
\frac{3}{2} \sin^4 x - \frac{4}{3} \sin^6 x + C
\end{aligned}
$$
which matches your original answer.