Question

How to integrate $\sin \left( {1 - 2x} \right)\cos \left( {1 + 2x} \right)$?

Answer 1

Hint: First, use trigonometry identity (II) to find the value of $\sin \left( {1 - 2x} \right)\cos \left( {1 + 2x} \right)$ and put the value of $\sin \left( {1 - 2x} \right)\cos \left( {1 + 2x} \right)$ in given integral. Then, use distributive property in integral. Next, use property (III) to split the integrals. The, solve the integral using integration formula (IV). Substitute all the values in the combined integral and get the required result.

Formula used:
1. The integral of the product of a constant and a function = the constant $ \times $ integral of the function.
i.e., $\int {\left( {kf\left( x \right)dx} \right)} = k\int {f\left( x \right)dx} $, where $k$ is a constant.
2. Trigonometric identity: $\sin A\cos B = \dfrac{1}{2}\left[ {\sin \left( {A + B} \right) + \sin \left( {A - B} \right)} \right]$
3. The integral of the sum or difference of a finite number of functions is equal to the sum or difference of the integrals of the various functions.
i.e., $\int {\left[ {f\left( x \right) \pm g\left( x \right)} \right]dx} = \int {f\left( x \right)dx} \pm \int {g\left( x \right)dx} $
4. Integration formula: $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C,n \ne - 1$ and \[\int {\sin xdx} = - \cos x + C\]


Complete step by step solution:
We have to find $\int {\sin \left( {1 - 2x} \right)\cos \left( {1 + 2x} \right)dx} $…(i)
Use identity $\sin A\cos B = \dfrac{1}{2}\left[ {\sin \left( {A + B} \right) + \sin \left( {A - B} \right)} \right]$, by putting $A = 1 - 2x$ and $B = 1 + 2x$.
$\sin \left( {1 - 2x} \right)\cos \left( {1 + 2x} \right) = \dfrac{1}{2}\left[ {\sin \left( {1 - 2x + 1 + 2x} \right) + \sin \left( {1 - 2x - 1 - 2x} \right)} \right]$
Add the terms in the bracket.
$ \Rightarrow \sin \left( {1 - 2x} \right)\cos \left( {1 + 2x} \right) = \dfrac{1}{2}\left[ {\sin \left( 2 \right) + \sin \left( { - 4x} \right)} \right]$
Use identity $\sin \left( { - x} \right) = - \sin x$ in above equation.
$ \Rightarrow \sin \left( {1 - 2x} \right)\cos \left( {1 + 2x} \right) = \dfrac{1}{2}\left[ {\sin \left( 2 \right) - \sin \left( {4x} \right)} \right]$
Now, put the value of $\sin \left( {1 - 2x} \right)\cos \left( {1 + 2x} \right)$ in integral (i).
$\int {\sin \left( {1 - 2x} \right)\cos \left( {1 + 2x} \right)dx} = \int {\dfrac{1}{2}\left[ {\sin \left( 2 \right) - \sin \left( {4x} \right)} \right]dx} $…(ii)
Use distributive property in integral (ii).
$\int {\sin \left( {1 - 2x} \right)\cos \left( {1 + 2x} \right)dx} = \int {\left[ {\dfrac{1}{2}\sin \left( 2 \right) - \dfrac{1}{2}\sin \left( {4x} \right)} \right]dx} $…(iii)
Now, using the property that the integral of the sum or difference of a finite number of functions is equal to the sum or difference of the integrals of the various functions.
i.e., $\int {\left[ {f\left( x \right) \pm g\left( x \right)} \right]dx} = \int {f\left( x \right)dx} \pm \int {g\left( x \right)dx} $
So, in above integral (iii), we can use above property
\[\int {\sin \left( {1 - 2x} \right)\cos \left( {1 + 2x} \right)dx} = \int {\dfrac{1}{2}\sin \left( 2 \right)dx} - \int {\dfrac{1}{2}\sin \left( {4x} \right)dx} \]…(iv)
Now, using the property that the integral of the product of a constant and a function = the constant $ \times $ integral of the function.
i.e., $\int {\left( {kf\left( x \right)dx} \right)} = k\int {f\left( x \right)dx} $
So, in above integral (iv), we can use above property
\[\int {\sin \left( {1 - 2x} \right)\cos \left( {1 + 2x} \right)dx} = \dfrac{1}{2}\sin \left( 2 \right)\int {dx} - \dfrac{1}{2}\int {\sin \left( {4x} \right)dx} \]…(v)
Now, using the integration formula $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C,n \ne - 1$ and \[\int {\sin xdx} = - \cos x + C\] in integration (v), we get
\[\int {\sin \left( {1 - 2x} \right)\cos \left( {1 + 2x} \right)dx} = \dfrac{1}{2}\sin \left( 2 \right)x - \dfrac{1}{2} \times \dfrac{{\cos \left( {4x} \right)}}{4} + C\]
It can be written as \[\int {\sin \left( {1 - 2x} \right)\cos \left( {1 + 2x} \right)dx} = \dfrac{1}{8}\left[ {4\sin \left( 2 \right)x + \cos \left( {4x} \right)} \right] + C\].

Hence, \[\int {\sin \left( {1 - 2x} \right)\cos \left( {1 + 2x} \right)dx} = \dfrac{1}{8}\left[ {4\sin \left( 2 \right)x + \cos \left( {4x} \right)} \right] + C\].

Note: To evaluate integrals of the form $\int {\sin mx\cos nxdx} $, $\int {\sin mx\sin nxdx} $, $\int {\cos mx} \cos nxdx$ and $\int {\cos mx\sin nxdx} $, we use the following trigonometric identities:
$2\sin A\cos B = \sin \left( {A + B} \right) + \sin \left( {A - B} \right)$
$2\cos A\sin B = \sin \left( {A + B} \right) - \sin \left( {A - B} \right)$
$2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)$
$2\sin A\sin B = \cos \left( {A - B} \right) - \cos \left( {A + B} \right)$