Answer
Verified
470.7k+ views
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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Which are the Top 10 Largest Countries of the World?
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Write a letter to the principal requesting him to grant class 10 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE