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 = $\dfrac{1}{2}$[-Cos(A+B) + Cos(A-B)]
⟹SinA SinB = $\dfrac{1}{2}$[Cos(A-B) – Cos(A+B)]
We compare this with Sinx Sin2x, we get A = x and B = 2x
⟹Sinx Sin2x = $\dfrac{1}{2}$[Cos(x-2x) – Cos(x+2x)]
⟹Sinx Sin2x = $\dfrac{1}{2}$[Cos(-x) – Cos3x]
We know for the trigonometric function cosine, Cos(-x) = Cos x.
So Sinx Sin2x = $\dfrac{1}{2}$[Cosx – Cos3x]
Hence our equation $\int {{\text{sinx sin2x sin3x}}} {\text{ dx}}$= $\int {\dfrac{1}{2}\left( {{\text{cosx - cos3x}}} \right)\sin 3{\text{x dx}}}$
= $\dfrac{1}{2}\int {\left( {{\text{cosx - cos3x}}} \right)\sin 3{\text{x dx}}}$
= $\dfrac{1}{2}\int {\left( {{\text{cosx sin3x - cos3x sin3x}}} \right){\text{dx}}}$
= $\dfrac{1}{2}\left[ {\int {{\text{cosx sin3x dx - }}\int {{\text{cos3x sin3x}}} {\text{ dx}}} } \right]$ - (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)
⟹${\text{SinA CosB}}$ = $\dfrac{1}{2}\left[ {{\text{Sin(A + B) + Sin(A - B)}}} \right]$
Comparing this to the above equation we get, A = 3x and B = x
⟹Sin3x Cosx = $\dfrac{1}{2}$[Sin(x+3x) + Sin(3x-x)]
= $\dfrac{1}{2}$[Sin4x + Sin2x]
Hence, $\int {{\text{Sin3x Cosx dx = }}\dfrac{1}{2}\int {\left[ {{\text{Sin4x + Sin2x}}} \right]} {\text{ dx}}}$.

Now let us solve $\int {{\text{cos3x sin3x}}} {\text{ dx}}$
We know that 2 SinA CosB = Sin(A+B) + Sin(A-B)
⟹${\text{SinA CosB}}$ = $\dfrac{1}{2}\left[ {{\text{Sin(A + B) + Sin(A - B)}}} \right]$
Comparing this to the above equation we get, A = 3x and B = 3x
⟹Sin3x Cos3x = $\dfrac{1}{2}$[Sin(3x+3x) + Sin(3x-3x)]
= $\dfrac{1}{2}$[Sin6x + Sin0]
= $\dfrac{1}{2}$Sin6x dx
Hence, $\int {{\text{Sin3x Cos3x dx = }}\dfrac{1}{2}\int {\left[ {{\text{Sin6x}}} \right]} {\text{ dx}}}$.
Thus, our original equation becomes
$\int {{\text{sinx sin2x sin3x}}} {\text{ dx}}$= $\dfrac{1}{2}\left[ {\int {{\text{cosx sin3x dx - }}\int {{\text{cos3x sin3x}}} {\text{ dx}}} } \right]$
= $\dfrac{1}{2}\left[ {\dfrac{1}{2}\int {\left( {{\text{sin4x + sin2x}}} \right){\text{ dx - }}\dfrac{1}{2}\int {\left( {{\text{sin6x}}} \right)} {\text{ dx}}} } \right]$
= $\dfrac{1}{4}\left[ {\int {{\text{sin4x dx + }}\int {{\text{sin2x dx}}} {\text{ - }}\int {{\text{sin6x dx}}} } } \right]$
We know$\int {{\text{sin}}\left( {{\text{ax + b}}} \right)} {\text{ dx = - }}\dfrac{{{\text{cos}}\left( {{\text{ax + b}}} \right)}}{{\text{a}}} + {\text{C}}$, 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,
$\int {{\text{sinx sin2x sin3x}}} {\text{ dx}}$= $\dfrac{1}{4}\left[ {\dfrac{{ - {\text{cos4x}}}}{4} + \left( {\dfrac{{ - {\text{cos2x}}}}{2}} \right) - \left( {\dfrac{{ - {\text{cos6x}}}}{6}} \right)} \right] + {\text{C}}$
= $\dfrac{1}{4}\left[ {\dfrac{{{\text{cos6x}}}}{6} - \dfrac{{{\text{cos4x}}}}{4} - \dfrac{{{\text{cos2x}}}}{2}} \right] + {\text{C}}$
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.