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**Hint**: We first discuss the integration by parts method. Integration by parts method is usually used for the multiplication of the functions and their integration. We take two arbitrary functions to express the theorem. We take the $ u=x,v=\sin x $ for our integration $ \int{x\sin xdx} $ . We use the formulas $ \int{\sin xdx}=-\cos x,\int{\cos xdx}=\sin x $ , $ \dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}} $ .

**:**

__Complete step-by-step answer__We need to find the integration of $ \int{x\sin xdx} $ using integration by parts method.

Integration by parts method is usually used for the multiplication of the functions and their integration.

Let’s assume $

f\left( x \right)=g\left( x \right)h\left( x \right) $ .

We need to find the integration of

$ \int{f\left( x \right)dx}=\int{g\left( x \right)h\left( x \right)dx} $ .

We take

$ u=g\left( x \right),v=h\left( x \right) $ . This gives $ \int{f\left( x \right)dx}=\int{uvdx} $ .

The theorem of integration by parts gives

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

For our integration $ \int{x\sin xdx} $ , we take $ u=x,v=\sin x $ .

Now we complete the integration

\[\int{x\sin xdx}=x\int{\sin xdx}-\int{\left( \dfrac{d\left( x \right)}{dx}\int{\sin xdx} \right)dx}\].

We have the differentiation formula for $ u=x $ where

$ \dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}} $ .

The integration formula for

$ \int{\sin xdx}=-\cos x,\int{\cos xdx}=\sin x $ .

We apply these formulas to complete the integration and get

\[\int{x\sin xdx}=x\left( -\cos x \right)-\int{\left( -\cos x \right)dx}=-x\cos x+\int{\cos xdx}\].

We have one more integration part remaining.

So, \[\int{x\sin xdx}=-x\cos x+\int{\cos xdx}=-x\cos x+\sin x+c\]. Here $ c $ is the integral constant.

Therefore, the integration by parts of $ \int{x\sin xdx} $ gives \[-x\cos x+\sin x+c\].

**So, the correct answer is “ \[-x\cos x+\sin x+c\]”.**

**Note**: In case one of two functions are missing and we need to form the by parts method, we will take the multiplying constant 1 as the second function.

For example: if we need to find \[\int{\ln xdx}\], we have only one function. So, we take constant 1 as the second function where $ u=\ln x,v=1 $ . But we need to remember that we won’t perform by parts by taking $ u=1 $ .

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