Find the integrals of the function: \(cos\, 2x\, cos\, 4x\, cos\, 6x\)
Solution and Explanation
It is known that, \(cosA\,cosB = \frac{1}{2}[cos(A+B)+cos(A-B)]\)
\(∴ ∫cos2x(cos4x\,cos6x)dx = ∫cos2x[\frac{1}{2}{cos(4x+6x)+cos(4x-6x)}]dx\)
\(=\frac{1}{2} ∫ {cos2x\,cos10x+cos2x\,cos(-2x)}dx\)
\(=\frac{1}{2} ∫[cos2x\,cos10x+cos^2 2x]dx\)
\(=\frac{1}{2} ∫[{\frac{1}{2}cos(2x+10x)+cos(2x-10x)}+(\frac{1+cos4x}{2})]dx\)
\(=\frac{1}{4} ∫(cos12x+cos8x+1+cos4x)dx\)
\(=\frac{1}{4}[\frac{sin12x}{12}+\frac{sin8x}{8}+x+\frac{sin4x}{4}]+C\)
Top Questions on integral
- Let $f : (0, \infty) \to \mathbb{R}$ and $F(x) = \int_{0}^{x} tf(t) \, dt$. If $F(x^2) = x^4 + x^5$, then \[\sum_{r=1}^{12} f(r^2)\]is equal to:
- If \[\int_{0}^{\frac{\pi}{3}} \cos^4 x \, dx = a\pi + b\sqrt{3},\]where $a$ and $b$ are rational numbers, then $9a + 8b$ is equal to:
- The value of \[\int_{0}^{1} \left(2x^3 - 3x^2 - x + 1\right)^{\frac{1}{3}} \, dx\]is equal to:
- Let \( r_k = \frac{\int_{0}^{1} (1 - x^7)^k \, dx}{\int_{0}^{1} (1 - x^7)^{k+1} \, dx}, \, k \in \mathbb{N} \). Then the value of \[ \sum_{k=1}^{10} \frac{1}{7(r_k - 1)}\]is equal to ______.
- \( \int_{0}^{\pi/4} \frac{\cos^2 x \sin^2 x}{\left( \cos^3 x + \sin^3 x \right)^2} \, dx \) is equal to:
Questions Asked in CBSE CLASS XII exam
- Find the inverse of each of the matrices,if it exists \(\begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}\)
- For what values of x,\(\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\)\(\begin{bmatrix} 1 & 2 & 0\\ 2 & 0 & 1 \\1&0&2 \end{bmatrix}\)\(\begin{bmatrix} 0 \\2\\x\end{bmatrix}\)=O?
What is the Planning Process?
- CBSE CLASS XII - 2023
- Planning process steps
- Find the inverse of each of the matrices,if it exists. \(\begin{bmatrix} 2 & 3\\ 5 & 7 \end{bmatrix}\)
- Find the inverse of each of the matrices, if it exists. \(\begin{bmatrix} 1 & 3\\ 2 & 7\end{bmatrix}\)
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
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