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$\int{x\sin 2xdx}$

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- Hint: To solve this problem, we will use the property of integration by parts to find the answer to the above integral. The formula for integration by parts is given by –

$\int{uvdx=u\int{vdx-\int{\dfrac{du}{dx}\left( \int{vdx} \right)}}}dx$

__Complete step-by-step solution__ -

Here, u and v are two functions of x. In our case, u = x and v = sin2x. We will use this to approach this problem.

To solve this problem, we have a combination of algebraic and trigonometric terms together. Thus, in such cases, we have to use integration by parts. For this, we will use the formula given by –$\int{uvdx=u\int{vdx-\int{\dfrac{du}{dx}\left( \int{vdx} \right)}}}dx$

In this case, u = x and v = sin2x, thus, we have,

L = $\int{x\sin 2xdx}$

L = $x\int{\sin 2xdx-\int{\dfrac{d(x)}{dx}\left( \int{\sin 2xdx} \right)}}dx$

We know that $\int{\sin nx dx=-\dfrac{\cos nx}{n}}$ and $\dfrac{d({{x}^{n}})}{dx}=n{{x}^{n-1}}$, thus, applying this formula here, we have,

L = $x\left( -\dfrac{\cos 2x}{2} \right)-\int{\left( -\dfrac{\cos 2x}{2} \right)}dx$

L = $-x\left( \dfrac{\cos 2x}{2} \right)+\int{\left( \dfrac{\cos 2x}{2} \right)}dx$

We will now use the formula, $\int{\cos nxdx=\dfrac{\sin nx}{n}}$

L = $-x\left( \dfrac{\cos 2x}{2} \right)+\dfrac{1}{2}\left( \dfrac{\sin 2x}{2} \right)$

L = $-x\left( \dfrac{\cos 2x}{2} \right)+\dfrac{1}{4}\left( \sin 2x \right)$

Further, since this is an indefinite integral, we need to have an integration constant to the final answer. Hence, the correct answer is $-x\left( \dfrac{\cos 2x}{2} \right)+\dfrac{1}{4}\left( \sin 2x \right)$ + c. Here, c is a constant.

Note: In general, if we have to solve the integration of the form $\int{{{x}^{n}}\sin nx dx}$, we have to perform the integration by part n number of times. This can be observed in the solution as by performing integration by parts, we would have ${{x}^{n}}\left( -\dfrac{\cos nx}{n} \right)-\int{n{{x}^{n-1}}\left( -\dfrac{\cos nx}{n} \right)}dx$. Thus, we can see that we have to differentiate the term ${{x}^{n}}$, n number of times for x to disappear. Further, we have to remember to add the integration constant at the end since integration is basically the reverse of differentiation. Thus, since, the differentiation of a constant term is 0 and such any answer say $-x\left( \dfrac{\cos 2x}{2} \right)+\dfrac{1}{4}\left( \sin 2x \right)$ + 2, $-x\left( \dfrac{\cos 2x}{2} \right)+\dfrac{1}{4}\left( \sin 2x \right)$ + 5 or any other term would be correct. Thus, we represent the end term by a general constant term, c.

$\int{uvdx=u\int{vdx-\int{\dfrac{du}{dx}\left( \int{vdx} \right)}}}dx$

Here, u and v are two functions of x. In our case, u = x and v = sin2x. We will use this to approach this problem.

To solve this problem, we have a combination of algebraic and trigonometric terms together. Thus, in such cases, we have to use integration by parts. For this, we will use the formula given by –$\int{uvdx=u\int{vdx-\int{\dfrac{du}{dx}\left( \int{vdx} \right)}}}dx$

In this case, u = x and v = sin2x, thus, we have,

L = $\int{x\sin 2xdx}$

L = $x\int{\sin 2xdx-\int{\dfrac{d(x)}{dx}\left( \int{\sin 2xdx} \right)}}dx$

We know that $\int{\sin nx dx=-\dfrac{\cos nx}{n}}$ and $\dfrac{d({{x}^{n}})}{dx}=n{{x}^{n-1}}$, thus, applying this formula here, we have,

L = $x\left( -\dfrac{\cos 2x}{2} \right)-\int{\left( -\dfrac{\cos 2x}{2} \right)}dx$

L = $-x\left( \dfrac{\cos 2x}{2} \right)+\int{\left( \dfrac{\cos 2x}{2} \right)}dx$

We will now use the formula, $\int{\cos nxdx=\dfrac{\sin nx}{n}}$

L = $-x\left( \dfrac{\cos 2x}{2} \right)+\dfrac{1}{2}\left( \dfrac{\sin 2x}{2} \right)$

L = $-x\left( \dfrac{\cos 2x}{2} \right)+\dfrac{1}{4}\left( \sin 2x \right)$

Further, since this is an indefinite integral, we need to have an integration constant to the final answer. Hence, the correct answer is $-x\left( \dfrac{\cos 2x}{2} \right)+\dfrac{1}{4}\left( \sin 2x \right)$ + c. Here, c is a constant.

Note: In general, if we have to solve the integration of the form $\int{{{x}^{n}}\sin nx dx}$, we have to perform the integration by part n number of times. This can be observed in the solution as by performing integration by parts, we would have ${{x}^{n}}\left( -\dfrac{\cos nx}{n} \right)-\int{n{{x}^{n-1}}\left( -\dfrac{\cos nx}{n} \right)}dx$. Thus, we can see that we have to differentiate the term ${{x}^{n}}$, n number of times for x to disappear. Further, we have to remember to add the integration constant at the end since integration is basically the reverse of differentiation. Thus, since, the differentiation of a constant term is 0 and such any answer say $-x\left( \dfrac{\cos 2x}{2} \right)+\dfrac{1}{4}\left( \sin 2x \right)$ + 2, $-x\left( \dfrac{\cos 2x}{2} \right)+\dfrac{1}{4}\left( \sin 2x \right)$ + 5 or any other term would be correct. Thus, we represent the end term by a general constant term, c.