Question

How do you evaluate the integral of $\int { - 9x\cos \left( {7x} \right)dx} $ ?

Answer 1

Hint: In this question, we are going to evaluate the integral.
In this, we are going to evaluate the integral by using integration by parts method. First, denote the given integral as ${\rm I}$ and then choose $u$ and $v$ from the given integral in the order LIATE.
We can differentiate and integrate wherever necessary by applying the integration by parts formula, we can get the required solution.

Formula used: Integration by parts
$\int {u \cdot dv = u \cdot v - \int {v \cdot du} } $
Choose $u$ in this order: LIATE
L - Logarithmic functions
I – Inverse trigonometric functions
A – Algebraic function
T – Trigonometric functions
E – Exponential functions
It is a rule which helps to decide which term you should differentiate first and which term should you integrate first. The term which is closer to L is differentiated first and the term which is closer to E is integrated first.

Complete step-by-step solution:
First let us take the given integral as follows:
${\rm I} = \int { - 9x\cos \left( {7x} \right)dx} $
Here the integral is denoted by ${\rm I}$
From this integral take $u$ and $dv$ as follows
We can choose $u$ as $ - 9x$ because it is an algebraic function and we should differentiate it.
$u = - 9x$
$\dfrac{{du}}{{dx}} = - 9$
We can choose $dv$ as $\cos \left( {7x} \right)dx$ because it is a trigonometric function and we should integrate it.
$\Rightarrow$$dv = \cos 7xdx$
$\Rightarrow$$\dfrac{{dv}}{{dx}} = \cos 7x$
$\Rightarrow$$\int {dv} = \int {\cos } 7xdx$
$\Rightarrow$$v = \dfrac{1}{7}\sin 7x$
Using integration by parts,
$\Rightarrow$${\rm I} = \int {u \cdot dv = u \cdot v - \int {v \cdot du} } $
Substitute $u,v,du$ in the above formula we get,
$\Rightarrow$${\rm I} = \left( { - 9x} \right) \cdot \dfrac{1}{7}\sin 7x - \int {\dfrac{1}{7}\sin 7x \cdot \left( { - 9} \right)} dx$
Solving the second part,
$\Rightarrow$${\rm I} = \dfrac{{ - 9}}{7}x\sin 7x + \int {\dfrac{9}{7}\sin 7x} dx$
$\Rightarrow$${\rm I} = \dfrac{{ - 9}}{7}x\sin 7x - \dfrac{9}{7}\left( {\dfrac{1}{7}\cos 7x} \right) + c$
Simplifying further,
$\Rightarrow$${\rm I} = \dfrac{{ - 9}}{7}x\sin 7x - \left( {\dfrac{9}{{49}}\cos 7x} \right) + c$
Thus, the integral of $\int { - 9x\cos \left( {7x} \right)dx} $ is $\dfrac{{ - 9}}{7}x\sin 7x - \left( {\dfrac{9}{{49}}\cos 7x} \right) + c$

Thus, $\dfrac{{ - 9}}{7}x\sin 7x - \left( {\dfrac{9}{{49}}\cos 7x} \right) + c$ is the required result.

Note: Formula used in this question is,
$\dfrac{d}{{dx}}nx = n$
Let us integrate $\cos \left( {ax} \right),$ where $a$ a constant is, and is not equal to zero.
$\int {\cos \left( {ax} \right)dx} $
Let us take $u = ax$
$du = adx$
$\dfrac{{du}}{{dx}} = a$
$\dfrac{{du}}{a} = dx$
$\Rightarrow$$\int {\cos \left( {ax} \right)dx = \int {\cos \left( u \right)\dfrac{{du}}{a}} } $
$\Rightarrow$$\int {\cos \left( {ax} \right)dx = \dfrac{1}{a}\int {\cos \left( u \right)du} } $
$\Rightarrow$$\int {\cos \left( {ax} \right)dx = \dfrac{1}{a}\sin u + c} \left[ {\because \int {\cos xdx = \sin x} } \right]$
$\Rightarrow$$\int {\cos \left( {ax} \right)dx = \dfrac{1}{a}\sin \left( {ax} \right) + c} $
So, for $\cos (7x)$ , we have
$\Rightarrow$$\int {\cos \left( {7x} \right)dx = \dfrac{1}{7}\sin \left( {7x} \right) + c} $
Similarly, we can get,
$\Rightarrow$$\int {\sin \left( {7x} \right)dx = - \dfrac{1}{7}\cos \left( {7x} \right) + c} $
Integration is a method of adding values on a large scale, where we cannot perform general addition operation. There are different integration methods that are used to find an integral of some function, which is easier to evaluate the original integral. We can evaluate the integral by using various methods of integration. They are
Integration by substitution, integration by parts, integration using trigonometric identities, integration of some particular function, integration by partial fraction.