Question

Evaluate $\int {x\cos (2x)dx} $ by integration by parts method ?

Answer 1

Hint:
In such questions where we are required to find the antiderivative or the integral of a function that is a product of two or more expressions or functions, we can make use of Integration by parts. Integration by parts method divides the product functions into halves where we differentiate the first half and integrate the second half.

Complete step by step solution:
In the given question, we are required to find the value of the integral given to us in the question by using the integration by parts method. So, we consider the given integral as a new variable.
Consider $I = \int {x\cos (2x)dx} $
In integration by parts method, we integrate a function which is a product of two functions using a formula:
$\int {f\left( x \right)g\left( x \right)dx = f\left( x \right)\int {g\left( x \right)dx - \int {\left[ {\dfrac{d}{{dx}}f\left( x \right)\int {g\left( x \right)dx} } \right]dx} } } $
Hence, Using integration by parts method and considering $x$as first function and $\cos (2x)$ as second function, we get
\[I = \left[ {x\int {\cos (2x)dx} } \right] - \int {\left[ {\dfrac{d}{{dx}}(x).\int {\cos (2x)dx} } \right]} dx\]
Now, we know that the derivative of x with respect to x is $1$. Also, we know that integral of $\cos x$ with respect to x is $\sin x$ and integral of $\cos 2x$ with respect to x is $\left( {\dfrac{{\sin 2x}}{2}} \right)$.
$I = \left[ {\dfrac{{x\sin (2x)}}{2}} \right] - \int {\left[ {\dfrac{{1.\sin (2x)}}{2}} \right]} dx$
Now, we know that the integral of $\sin x$with respect to x is $ - \cos x$ and integral of $\sin 2x$ is $\left( {\dfrac{{ - \cos 2x}}{2}} \right)$. Hence, substituting the values of the known derivatives, we get the value of original integral as:
$I = \left[ {\dfrac{{x\sin (2x)}}{2}} \right] - \left[ {\dfrac{{ - \cos (2x)}}{{2 \times 2}}} \right] + C$
Opening the brackets and simplifying,
$I = \left[ {\dfrac{{x\sin (2x)}}{2}} \right] + \left[ {\dfrac{{\cos (2x)}}{4}} \right] + C$
$I = \dfrac{1}{4}\left[ {2x\sin (2x) + \cos (2x)} \right] + C$

So, the value of integral $\int {x\cos (2x)dx} $ is $\dfrac{1}{4}\left[ {2x\sin (2x) + \cos (2x)} \right] + C$, where C is any arbitrary constant.

Note:
Integration by parts method can be used to solve integrals of various complex functions involving products of functions within itself easily. In the given question, the use of integration by parts method once has made the process much easier and organized. However, this might not be the case every time. We may have to apply the integration by parts method multiple times in order to reach the final answer.