Question

How do you integrate $ \int {{x^2}\sin (2x)dx} $ from $ \left[ {0,\dfrac{\pi }{2}} \right] $ ?

Answer 1

Hint: We can see that there are two different functions in the given integral. Therefore, we have to apply the rule of integration by part to solve it. We will first solve the integral and then apply the limits to get our final answer.
Formula used:
 $ \int {uvdx = u\int {vdx - \int {u'\left( {\int {vdx} } \right)} } } dx $

Complete step-by-step answer:
Let us take $ I = \int {{x^2}\sin (2x)dx} $ .
Now, we have to determine $ u $ and $ v $ functions between the two given functions.
Therefore, let us take $ u = {x^2} $ and $ v = \sin 2x $ .
 $ u = {x^2} \Rightarrow u' = 2x $ and $ v = \sin 2x \Rightarrow \int {vdx = \int {\sin 2xdx = - \dfrac{1}{2}\cos 2x} } $ .
Now, we will use these values in the formula for integration by parts:
 $ \int {uvdx = u\int {vdx - \int {u'\left( {\int {vdx} } \right)} } } dx $
 \[
  \int {{x^2}\sin (2x)dx} \\
   = {x^2}\int {\sin (2x)dx - \int {\dfrac{d}{{dx}}} } {x^2}\left( {\int {\sin (2x)dx} } \right)dx \\
   = {x^2} \times - \dfrac{1}{2}\cos 2x - \int {\left( {2x \times - \dfrac{1}{2}\cos 2x} \right)} dx \\
   = - \dfrac{{{x^2}}}{2}\cos 2x + \int {xcox2xdx} \\
 \]
Now, we have \[I = - \dfrac{{{x^2}}}{2}\cos 2x + \int {xcox2xdx} \] .
We can see that there are still two different functions in the second term. Therefore, to solve that, we will again need to apply the formula for integration by parts.
We have to determine $ u $ and $ v $ functions between the two given functions in \[\int {xcox2xdx} \] .
Therefore, let us take $ u = x $ and $ v = \cos 2x $ .
 $ u = x \Rightarrow u' = 1 $ and $ v = \cos 2x \Rightarrow \int {vdx = \int {\cos 2xdx = \dfrac{1}{2}\sin 2x} } $ .
Now, we will use these values in the formula for integration by parts:
 $ \int {uvdx = u\int {vdx - \int {u'\left( {\int {vdx} } \right)} } } dx $
 \[
  \int {x\cos 2xdx} \\
   = x\int {\cos 2xdx - \int {\dfrac{d}{{dx}}} } {x^2}\left( {\int {\cos 2xdx} } \right)dx \\
   = x \times \dfrac{1}{2}\sin 2x - \int {\left( {1 \times \dfrac{1}{2}\sin 2x} \right)} dx \\
   = \dfrac{x}{2}\sin 2x + \dfrac{1}{4}\cos 2x \\
 \]
Now, we have \[I = - \dfrac{{{x^2}}}{2}\cos 2x + \dfrac{x}{2}\sin 2x + \dfrac{1}{4}\cos 2x\] .
Now, we need to find the integral in the interval of $ \left[ {0,\dfrac{\pi }{2}} \right] $ .
Therefore we can write it as:
 \[
  \int\limits_0^{\dfrac{\pi }{2}} {{x^2}\sin (2x)} dx \\
   = \left[ { - \dfrac{{{x^2}}}{2}\cos 2x + \dfrac{x}{2}\sin 2x + \dfrac{1}{4}\cos 2x} \right] _0^{\dfrac{\pi }{2}} \\
   = \left[ {\dfrac{{{\pi ^2}}}{8} - \dfrac{1}{4}} \right] - \left[ {\dfrac{1}{4}} \right] \\
   = \dfrac{{{\pi ^2}}}{8} - \dfrac{1}{2} \\
   = 0.73 \;
 \]
Thus, our final answer is $ 0.73 $ .
So, the correct answer is “ $ 0.73 $ ”.

Note: Here, we have applied the formula for integration by parts two times while solving the integral. In this rule, it is important to determine which function to take as $ u $ and which one is to take as $ v $ . For this, we need to keep in mind that we should choose the function $ u $ that gets simpler when we differentiate it and the function $ v $ that does not get any more complicated when you integrate it.