Question:

Find the integrals of the function: \(sin\, x\,\, sin\, 2x\,\, sin\, 3x\)

Updated On: Oct 19, 2023
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Solution and Explanation

The correct answer is: \(= \frac{1}{8}[\frac{cos6x}{3}-\frac{cos4x}{2}-cos2x]+C\)
It is known that, \(sin A\, sin B = \frac{1}{2}{cos(A-B)-cos(A+B)}\)
\(∴ ∫sin\, x\,\, sin\, 2x\,\, sin\, 3x. dx = ∫[sin x.\frac{1}{2}{cos(2x-3x)-cos(2x+3x)}]dx\)
\(= \frac{1}{2}∫(sinx\, cos(-x)-sinx\, cos\,5x)dx\)
\(= \frac{1}{2} ∫(sinx\,cosx-sinx\,cos5x)dx\)
\(= \frac{1}{2} ∫\frac{sin2x}{2}. dx -\frac{1}{2} ∫sin x\, cos5x. dx\)
\(= \frac{1}{4}[\frac{-cos2x}{2}]-\frac{1}{2} ∫{\frac{1}{2}sin(x+5x)+sin(x-5x)}dx\)
\(= \frac{-cos2x}{8}-\frac{1}{4} ∫(sin6x+sin(-4x))dx\)
\(=\frac{-cos2x}{8}-\frac{1}{4}[\frac{-cos6x}{3}+\frac{cos4x}{4}]+C\)
\(=\frac{-cos2x}{8}-\frac{1}{8}[\frac{-cos6x}{3}+\frac{cos4x}{2}]+C\)
\(= \frac{1}{8}[\frac{cos6x}{3}-\frac{cos4x}{2}-cos2x]+C\)
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Concepts Used:

Methods of Integration

Given below is the list of the different methods of integration that are useful in simplifying integration problems:

Integration by Parts:

 If f(x) and g(x) are two functions and their product is to be integrated, then the formula to integrate f(x).g(x) using by parts method is:

∫f(x).g(x) dx = f(x) ∫g(x) dx − ∫(f′(x) [ ∫g(x) dx)]dx + C

Here f(x) is the first function and g(x) is the second function.

Method of Integration Using Partial Fractions:

The formula to integrate rational functions of the form f(x)/g(x) is:

∫[f(x)/g(x)]dx = ∫[p(x)/q(x)]dx + ∫[r(x)/s(x)]dx

where

f(x)/g(x) = p(x)/q(x) + r(x)/s(x) and

g(x) = q(x).s(x)

Integration by Substitution Method

Hence the formula for integration using the substitution method becomes:

∫g(f(x)) dx = ∫g(u)/h(u) du

Integration by Decomposition

Reverse Chain Rule

This method of integration is used when the integration is of the form ∫g'(f(x)) f'(x) dx. In this case, the integral is given by,

∫g'(f(x)) f'(x) dx = g(f(x)) + C

Integration Using Trigonometric Identities