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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?

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    $\begingroup$ It very well could be that both answers are correct, but differ by a constant. $\endgroup$ – Omnomnomnom Aug 11 '15 at 12:47
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    $\begingroup$ @Omnomnomnom Indeed that is what happened; Dominik's answer links to a Wolfram Alpha calculation showing the two differ by $7/48$. $\endgroup$ – David K Aug 11 '15 at 13:00
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    $\begingroup$ So they're both right. Dang trig identities. $\endgroup$ – Akiva Weinberger Aug 11 '15 at 13:45
  • $\begingroup$ Thank you for having legible handwriting. $\endgroup$ – iamnotmaynard Aug 11 '15 at 14:57
  • $\begingroup$ You kept everything based on sines, while your textbook converted to cosines. You might want to check the assignment more closely to see if you didn't accidentally miss a detail that states you need to write everything as a cosine, or something in the chapter even. $\endgroup$ – Nzall Aug 11 '15 at 22:14
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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: $f(x) = 1 + g(x)$.

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$ \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.

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$\bf{My\; Solution::}$ 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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  • $\begingroup$ how my solution is different from yours? $\endgroup$ – Yogesh Tripathi Aug 11 '15 at 13:25

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