Question

How do you integrate \[\int{x\sin x}\] by integration by parts method?

Answer 1

Hint: From the question given, we have been asked to integrate \[\int{x\sin x}\] by integration by parts method. To solve the given question by using the integration by parts formula, first of all, we have to know the formula for integration by parts. If we are studying integration, then we should have learnt the formula for integration by parts and how to use it. The formula for integration by parts is given as \[\int{u\dfrac{dv}{dx}dx=uv-\int{v\dfrac{du}{dx}}}\]

Complete step by step answer:
From the given question, we have been given that \[\int{x\sin x}\]
Let \[u=x\]
Then, derivative for it is shown below \[\dfrac{du}{dx}=1\]
And
\[\dfrac{dv}{dx}=\sin x\]
\[\Rightarrow v=-\cos x\]
Now, we have got all the values which we need to substitute in the integration by parts formula.
Therefore substitute the values we got in the integration by parts formula to solve the given question.
By substituting all the values we got above in the integration by parts formula, we get
\[\int{u\dfrac{dv}{dx}dx=uv-\int{v\dfrac{du}{dx}}}\]
\[\Rightarrow \int{x\sin xdx=\left( x \right)\left( -\cos x \right)-\int{\left( -\cos x \right)\left( 1 \right)dx}}\]
Now, simplify further to get the exact answer.
By simplifying furthermore, we get
\[\Rightarrow \int{x\sin xdx=-x\cos x+\int{\cos xdx}}\]
\[\Rightarrow \int{x\sin xdx=-x\cos x+\sin x+c}\]
Hence, the given question is solved by using the integration by parts formula.

Note: We should be well known about the integration by parts formula. Also, we should be well known about the application of integration by parts formula. Also, we should be very careful while doing the calculation part of the problem. Also, we should be very careful while applying the integration by parts formula. Similarly we have many more integration formulae like $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1},n\ne -1}$ and many more.