Question

# Find $\int\limits_{0}^{{\pi }/{2}\;}{x\sin xdx}$

Hint: Use the integration by parts formula,
$\int{u(x)\cdot v(x)dx}=u(x)\cdot \int{v(x)dx}-\int{\left( \dfrac{d}{dx}u(x)\cdot \int{v(x)dx} \right)dx}$ with $u(x)=x$ and $v(x)=\sin x$. and substitute upper and lower limit values.

First, let us decompose the function that we have to find the integration of (the integrand) into two functions.
Thus, $\int\limits_{0}^{{\pi }/{2}\;}{x\sin xdx}=\int\limits_{0}^{{\pi }/{2}\;}{\left( x \right)\cdot \left( \sin x \right)dx}$
We do this so that now we have two different functions and each of these functions is a simple function. Now, since we have 2 different functions, we can use integration by parts to solve the required integration. Using the ILATE rule for integration by parts, we can apply the formula$\int{u(x)\cdot v(x)dx}=u(x)\cdot \int{v(x)dx}-\int{\left( \dfrac{d}{dx}u(x)\cdot \int{v(x)dx} \right)dx}$.
In the above formula, by the ILATE rule, the function $u(x)=x$ and $v(x)=\sin x$.
Using these in the formula,
$\int\limits_{0}^{{\pi }/{2}\;}{x\sin xdx}=x\cdot \int{\sin xdx}-\int{\left( \dfrac{d}{dx}\left( x \right)\cdot \int{\sin xdx} \right)dx}\ \ \ \ \ \ \ \ldots \left( 1 \right)$
We know that $\dfrac{d}{dx}\left( x \right)=1$ and$\int{\sin xdx}=-\cos x$
Using the values of $\dfrac{d}{dx}\left( x \right)=1$ and $\int{\sin xdx}=-\cos x$ in the equation (1)
\begin{align} & \int\limits_{0}^{{\pi }/{2}\;}{x\sin xdx}=x\cdot \int{\sin xdx}-\int{\left( \dfrac{d}{dx}\left( x \right)\cdot \int{\sin xdx} \right)dx} \\ & \Rightarrow \int\limits_{0}^{{\pi }/{2}\;}{x\sin xdx}=\left( x \right)\cdot \left( -\cos x \right)-\int{\left( 1\cdot \left( \cos x \right) \right)dx} \\ & \Rightarrow \int\limits_{0}^{{\pi }/{2}\;}{x\sin xdx}=-x\cos x-\int{\cos xdx}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 2 \right) \\ \end{align}
We know that $\int{\cos xdx}=\sin x$
Using the value of $\int{\cos xdx}=\sin x$ in equation (2),
$\int\limits_{0}^{{\pi }/{2}\;}{x\sin xdx}=-x\cos x-\sin x+C$
Putting in the lower and upper limit in the integration obtained,
\begin{align} & \int\limits_{0}^{{\pi }/{2}\;}{x\sin xdx}=\left( -x\cos x-\sin x+C \right)_{0}^{{\pi }/{2}\;} \\ & \Rightarrow \int\limits_{0}^{{\pi }/{2}\;}{x\sin xdx}=\left( -\dfrac{\pi }{2}\cos \left( \dfrac{\pi }{2} \right)-\sin \left( \dfrac{\pi }{2} \right) \right)-\left( 0\cos 0-\sin 0 \right)\ \ \ \ \ \ \ \ldots \left( 3 \right) \\ \end{align}
Now we know that the value of $\cos \dfrac{\pi }{2}=0,\ \cos 0=1,\ \sin \dfrac{\pi }{2}=1$and $\sin 0=0$
Using these values in equation 3 we get
\begin{align} & \int\limits_{0}^{{\pi }/{2}\;}{x\sin xdx}=\left( \left( -\dfrac{\pi }{2} \right)\cdot \left( 0 \right)-1 \right)-\left( \left( 0 \right)\cdot \left( 1 \right)-0 \right) \\ & \Rightarrow \int\limits_{0}^{{\pi }/{2}\;}{x\sin xdx}=\left( -1 \right)-0 \\ & \Rightarrow \int\limits_{0}^{{\pi }/{2}\;}{x\sin xdx}=-1 \\ \end{align}
Thus the required integration of $\int\limits_{0}^{{\pi }/{2}\;}{x\sin xdx}$ is -1.

Note: The question is a basic question of integration by parts, but it is important to do the calculations carefully. One can be easily confused in the calculations involved with lots of positive and negative signs. To keep things simple, first find the complete indefinite integral and then put in limits.