Find the integrals of the function: \(\frac{cos2 x-cos2 α}{cos x - cos α}\)
Solution and Explanation
\(\frac{cos2 x-cos2 α}{cos x - cos α}\)\(=\frac{ -2 sin (\frac{2x+2α}{2}) sin (\frac{2x-2α}{2}) }{-2sin (\frac{x+α}{2}) sin (\frac{x- α}{2}) }\) [cos C - cos D = -2sin \(\frac{C+D}{2}\) sin \(\frac{C-D}{2}\)]
\(=\frac{sin(x+α)sin(x-α) }{sin(\frac{x+α}{2} )sin(\frac{x-α}{2} )}\)
\(=\frac{[2sin(\frac{x+ α}{2}) )cos(\frac{x+ α}{2}) )][2sin(\frac{x- α}{2}) )cos(\frac{x- α}{2}) )]}{sin(\frac{x+ α}{2}) )sin(\frac{x- α}{2}) )}\)
\(=4 cos (\frac{x+ α}{2}) cos(\frac{x- α}{2}) \)
\(=2[cos(\frac{x+ α}{2} +\frac{x- α}{2}) +cos \frac{x+ α}{2} - \frac{x- α}{2} ]\)
\(=2[cos(x) +cos α]\)
\(=2 cos x+ 2cos α\)
∴ ∫\(\frac{cos2 x-cos2 α}{cos x - cos α}\) = \(∫2 cos x+ 2cos α\)
\(=2[sin x+ x cos α]+C\)
Top Questions on integral
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Questions Asked in CBSE CLASS XII exam
- Find the inverse of each of the matrices, if it exists. \(\begin{bmatrix} 1 & 3\\ 2 & 7\end{bmatrix}\)
- 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?
- Find the inverse of each of the matrices,if it exists \(\begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}\)
What is the Planning Process?
- CBSE CLASS XII - 2023
- Planning process steps
- Find the inverse of each of the matrices,if it exists. \(\begin{bmatrix} 2 & 3\\ 5 & 7 \end{bmatrix}\)
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