Question

Evaluate: \[\int_{ - x}^x {{{(\cos ax - \sin bx)}^{2dx}}} \] .

Answer 1

Hint: It is a question based on integration. Do you know what integration is? Integration is the calculation of an integral. It denotes the summation of discrete data, also you must already know that it can be said as the reverse process of differentiation. You are given a certain limit in the question, you will have to integrate the given function in that certain limit or range. Let’s do it.

Step wise solution:
Given data: To evaluate the value of \[\int_{ - x}^x {{{(\cos ax - \sin bx)}^{2dx}}} \] .
To simplify the given function.
Here, \[{(\cos ax - \sin bx)^2}\] is in the form of\[{(A - B)^2}\,\,where\,\,A = \cos Ax\,\,and\,\,B = \sin bx\]
We know that,
$
{(A - B)^2}\, = {A^2} + {B^2} - 2AB\\ $
Similarly
$ {(\cos ax - \sin bx)^2} = {\cos ^2}ax + {\sin ^2}bx - 2\cos ax\sin bx $
To integrate the expanded form of the function
Suppose,
$ I = \int_{ - x}^x {(\cos ax} - \sin bx{)^2}dx\\
 \Rightarrow I = \int_{ - x}^x {({{\cos }^2}ax} + {\sin ^2}bx - 2\cos ax\sin bx)dx $
Since we knew, \[\int {[f(x) + g(x) + h(x)]dx} \] can be computed as
$ \int {[f(x) + g(x) + h(x)]dx} = [\int {f(x)} + \int {g(x) + \int {h(x)} ]dx} \\
 \Rightarrow \int {[f(x) + g(x) + h(x)]dx} = \int {f(x)} dx + \int {g(x)dx + \int {h(x)} dx]} $
Now, I can be rewritten as,
$ I = \int_{ - x}^x {{{\cos }^2}ax\,} dx + \int_{ - x}^x {{{\sin }^2}bx\,} dx - \int_{ - x}^x {2\cos ax\sin bx\,} dx $
Here, you can see are even functions and is an odd function, So, I is given by
$ I = 2\int_0^x {{{\cos }^2}ax\,} dx + 2\int_0^x {{{\sin }^2}bx\,} dx - 0 $
Here, we can write $ {\cos ^2}ax\,\,as\,\,\dfrac{{\cos 2ax + 1}}{2} $ because
$ 2{\cos ^2}ax - 1 = \,\,\cos 2ax + 1\\
i.e.:{\cos ^2}ax\,\, = \,\dfrac{{\cos 2ax + 1}}{2}\\ $
and $ \,\\
\,1 - 2{\sin ^2}bx = \cos 2bx\\
 \Rightarrow - 2{\sin ^2}bx = \cos 2bx - 1\\
 \Rightarrow {\sin ^2}bx = - \left( {\dfrac{{\cos 2bx - 1}}{2}} \right)\\
 \Rightarrow {\sin ^2}bx = \dfrac{{1 - \cos 2bx}}{2}\\ $
So, $ \,\\
I = 2\int_0^x {\left( {\dfrac{{1 - \cos 2ax}}{2}} \right)} dx + 2\int_0^x {\left( {\dfrac{{1 - \cos 2bx}}{2}} \right)} dx\\
 \Rightarrow I = \dfrac{2}{2}\int_0^x {(1 + \cos 2ax)dx + \dfrac{2}{2}\int_0^x {(1 - \cos 2bx)dx} } \\
 \Rightarrow I = \int_0^x ( 1 + \cos 2ax)dx + \int_0^x {(1 - \cos 2bx)dx} \\
 \Rightarrow I = \int_0^x ( 1 + \cos 2ax + 1 - \cos 2bx)dx\\
 \Rightarrow I = \int_0^x ( 2 + \cos 2ax - \cos 2bx)dx\\ $
Again,$ \\
 \Rightarrow I = \int_0^x 2 \,dx + \int_0^x {\cos 2ax} \,dx - \int_0^x {\cos 2bx} \,dx\\
 \Rightarrow I = 2\left[ x \right]_0^x + \left[ {\dfrac{{\sin 2ax}}{{2a}}} \right]_0^x - \left[ {\dfrac{{\sin 2bx}}{{2b}}} \right]_0^x\\
 \Rightarrow I = \left( {2x - 0} \right) + \left( {\dfrac{{\sin 2ax}}{{2a}} - 0} \right) - \left( {\dfrac{{\sin 2bx}}{{2b}} - 0} \right)\\
 \Rightarrow I = 2x + \dfrac{{\sin 2ax}}{{2a}} - \dfrac{{\sin 2bx}}{{2b}}\\ $
Hence, $ \\
\int_{ - x}^x {{{(\cos ax - \sin bx)}^2}dx = } 2x + \dfrac{{\sin 2ax}}{{2a}} - \dfrac{{\sin 2bx}}{{2b}} $
is the required answer to be evaluated.

Note: Students often make mistakes in the formulae of integration. Try to make the functions as simple as possible before integrating them to avoid mess in integration.