Find the integrals of the function: \(cos\, 2x\, cos\, 4x\, cos\, 6x\)
Approach Solution - 1
Step 1: Use the identity \( \cos A \cos B = \frac{\cos(A+B) + \cos(A-B)}{2} \). First combine \( \cos 2x \) and \( \cos 4x \): \[ \cos 2x \cdot \cos 4x = \frac{\cos(6x) + \cos(-2x)}{2} = \frac{\cos 6x + \cos 2x}{2}. \]
Step 2: Multiply by \( \cos 6x \): \[ \cos 2x \, \cos 4x \, \cos 6x = \frac{1}{2} \left[ \cos 6x \cdot \cos 6x + \cos 2x \cdot \cos 6x \right]. \]
Step 3: Use \( \cos^2\theta = \frac{1+\cos 2\theta}{2} \) for the first term: \[ \cos 6x \cdot \cos 6x = \cos^2 6x = \frac{1 + \cos 12x}{2}. \] For the second term, again use the product-to-sum formula: \[ \cos 2x \cdot \cos 6x = \frac{\cos(8x) + \cos(-4x)}{2} = \frac{\cos 8x + \cos 4x}{2}. \]
Step 4: Combine results: \[ \cos 2x \cos 4x \cos 6x = \frac{1}{2} \left[ \frac{1 + \cos 12x}{2} + \frac{\cos 8x + \cos 4x}{2} \right]. \] Simplifying: \[ = \frac{1}{4} \left[ 1 + \cos 12x + \cos 8x + \cos 4x \right]. \]
Step 5: Integrate term by term: \[ \int \cos 2x \cos 4x \cos 6x \, dx = \frac{1}{4} \left[ \int 1 \, dx + \int \cos 12x \, dx + \int \cos 8x \, dx + \int \cos 4x \, dx \right]. \] \[ = \frac{1}{4} \left[ x + \frac{\sin 12x}{12} + \frac{\sin 8x}{8} + \frac{\sin 4x}{4} \right] + C. \]
Final Answer: \[ \boxed{ \int \cos 2x \cos 4x \cos 6x \, dx = \frac{x}{4} + \frac{\sin 12x}{48} + \frac{\sin 8x}{32} + \frac{\sin 4x}{16} + C } \]
Approach Solution -2
It is known that, \(cosA\,cosB = \frac{1}{2}[cos(A+B)+cos(A-B)]\)
\(∴ ∫cos2x(cos4x\,cos6x)dx = ∫cos2x[\frac{1}{2}{cos(4x+6x)+cos(4x-6x)}]dx\)
\(=\frac{1}{2} ∫ {cos2x\,cos10x+cos2x\,cos(-2x)}dx\)
\(=\frac{1}{2} ∫[cos2x\,cos10x+cos^2 2x]dx\)
\(=\frac{1}{2} ∫[{\frac{1}{2}cos(2x+10x)+cos(2x-10x)}+(\frac{1+cos4x}{2})]dx\)
\(=\frac{1}{4} ∫(cos12x+cos8x+1+cos4x)dx\)
\(=\frac{1}{4}[\frac{sin12x}{12}+\frac{sin8x}{8}+x+\frac{sin4x}{4}]+C\)
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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