Question

Find the integral of $\text{sinx}\text{.sin2x}\text{.sin3x}$, denoted by $\int \text{sinx sin2x sin3x}\text{ dx}$.

Answer 1

Hint – To find the integral of $\text{sinx}\text{.sin2x}\text{.sin3x}$, we simplify the term using trigonometric identities and then apply the formulas for integrals of sine and cosine functions to approach the answer.

Complete step by step answer:
Let us start off by simplifying the first two terms of the equation,
$\int \text{sinx sin2x sin3x}\text{ dx}$= $\int \text{(sinx sin2x) sin3x}\text{ dx}$

We know that 2 SinA SinB = - cos (A + B) + cos (A – B)
⟹SinA SinB = ${\displaystyle \frac{1}{2}}$[-Cos(A+B) + Cos(A-B)]
⟹SinA SinB = ${\displaystyle \frac{1}{2}}$[Cos(A-B) – Cos(A+B)]
We compare this with Sinx Sin2x, we get A = x and B = 2x
⟹Sinx Sin2x = ${\displaystyle \frac{1}{2}}$[Cos(x-2x) – Cos(x+2x)]
⟹Sinx Sin2x = ${\displaystyle \frac{1}{2}}$[Cos(-x) – Cos3x]
We know for the trigonometric function cosine, Cos(-x) = Cos x.
So Sinx Sin2x = ${\displaystyle \frac{1}{2}}$[Cosx – Cos3x]
Hence our equation $\int \text{sinx sin2x sin3x}\text{ dx}$= $\int {\displaystyle \frac{1}{2}}\left(\text{cosx - cos3x}\right)\sin 3\text{x dx}$
                                                                               = ${\displaystyle \frac{1}{2}}\int \left(\text{cosx - cos3x}\right)\sin 3\text{x dx}$
                                                                               = ${\displaystyle \frac{1}{2}}\int \left(\text{cosx sin3x - cos3x sin3x}\right)\text{dx}$
                                                                               = ${\displaystyle \frac{1}{2}}\left[\int \text{cosx sin3x dx - }\int \text{cos3x sin3x}\text{ dx}\right]$ - (1)
We solve both the terms in equation (1) individually for easy simplification,
First let us solve$\int \text{cosx sin3x dx}$:
We know that 2 SinA CosB = Sin(A+B) + Sin(A-B)
                      ⟹$\text{SinA CosB}$ = ${\displaystyle \frac{1}{2}}\left[\text{Sin(A + B) + Sin(A - B)}\right]$
Comparing this to the above equation we get, A = 3x and B = x
⟹Sin3x Cosx = ${\displaystyle \frac{1}{2}}$[Sin(x+3x) + Sin(3x-x)]
                         = ${\displaystyle \frac{1}{2}}$[Sin4x + Sin2x]
Hence, $\int \text{Sin3x Cosx dx = }{\displaystyle \frac{1}{2}}\int \left[\text{Sin4x + Sin2x}\right]\text{ dx}$.

Now let us solve $\int \text{cos3x sin3x}\text{ dx}$
We know that 2 SinA CosB = Sin(A+B) + Sin(A-B)
                      ⟹$\text{SinA CosB}$ = ${\displaystyle \frac{1}{2}}\left[\text{Sin(A + B) + Sin(A - B)}\right]$
Comparing this to the above equation we get, A = 3x and B = 3x
⟹Sin3x Cos3x = ${\displaystyle \frac{1}{2}}$[Sin(3x+3x) + Sin(3x-3x)]
                         = ${\displaystyle \frac{1}{2}}$[Sin6x + Sin0]
                         = ${\displaystyle \frac{1}{2}}$Sin6x dx
Hence, $\int \text{Sin3x Cos3x dx = }{\displaystyle \frac{1}{2}}\int \left[\text{Sin6x}\right]\text{ dx}$.
Thus, our original equation becomes
$\int \text{sinx sin2x sin3x}\text{ dx}$= ${\displaystyle \frac{1}{2}}\left[\int \text{cosx sin3x dx - }\int \text{cos3x sin3x}\text{ dx}\right]$
                                           = ${\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{1}{2}}\int \left(\text{sin4x + sin2x}\right)\text{ dx - }{\displaystyle \frac{1}{2}}\int \left(\text{sin6x}\right)\text{ dx}\right]$
                                           = ${\displaystyle \frac{1}{4}}\left[\int \text{sin4x dx + }\int \text{sin2x dx}\text{ - }\int \text{sin6x dx}\right]$
We know$\int \text{sin}\left(\text{ax + b}\right)\text{ dx = - }{\displaystyle \frac{\text{cos}\left(\text{ax + b}\right)}{\text{a}}}+\text{C}$, where C is the integral constant.
Comparing this formula with the above equation we get, a=4, a=2 and a=6 for first, second and third terms respectively and b=0 for all terms.
Hence our equation becomes,
$\int \text{sinx sin2x sin3x}\text{ dx}$= ${\displaystyle \frac{1}{4}}\left[{\displaystyle \frac{-\text{cos4x}}{4}}+\left({\displaystyle \frac{-\text{cos2x}}{2}}\right)-\left({\displaystyle \frac{-\text{cos6x}}{6}}\right)\right]+\text{C}$
                                          = ${\displaystyle \frac{1}{4}}\left[{\displaystyle \frac{\text{cos6x}}{6}}-{\displaystyle \frac{\text{cos4x}}{4}}-{\displaystyle \frac{\text{cos2x}}{2}}\right]+\text{C}$
Hence the answer.

Note – In order to solve this type of questions the key is to have a very good knowledge in trigonometric identities and formulas of trigonometric functions, which include the functions sin and cos, as they play a key role in simplification. It is also very important to split the terms and solve them individually as it is easy to approach the answer that way.