# Evaluate $\displaystyle \int_{0}^{\Large \frac{\pi}{2}}{x\cos{x}}\,dx$

The definite integral of the product of $x$ and cosine of angle $x$ should be evaluated with respect to $x$ from $0$ to $\pi$ divided by $2$ in this definite integration problem.

It is given that two functions $x$ and $\cos{x}$ are multiplied in this definite integral question. One function $x$ can be differentiated and the integral of other function $\cos{x}$ can be evaluated with respect to $x$. So, the integration by parts formula can be used to find the definite integral of the product of $x$ and $\cos{x}$ with respect to $x$ over the interval $\Big[0,\,\dfrac{\pi}{2}\Big]$.

### Find Derivative and Antiderivative of functions

Assume $u \,=\, x$ and $dv \,=\, \cos{x}dx$

Now, differentiate the equation $u \,=\, x$ with respect to $x$ to find the differential $du$.

$\implies$ $\dfrac{d}{dx}{u} \,=\, \dfrac{d}{dx}{x}$

$\implies$ $\dfrac{du}{dx} \,=\, \dfrac{dx}{dx}$

$\implies$ $\dfrac{du}{dx} \,=\, 1$

$\implies$ $du \,=\, 1 \times dx$

$\,\,\,\therefore\,\,\,\,\,\,$ $du \,=\, dx$

It is time to find the antiderivative of expressions in the differential equation $dv \,=\, \cos{x}\,dx$ by the indefinite integration.

$\implies$ $\displaystyle \int{}dv$ $\,=\,$ $\displaystyle \int{\cos{x}}\,dx$

$\implies$ $\displaystyle \int{1}\,dv$ $\,=\,$ $\displaystyle \int{\cos{x}}\,dx$

According to the integral of one rule, the indefinite integral of one can be evaluated with respect to $v$. Similarly, the indefinite integral of cosine of angle $x$ can also be evaluated by the integral of cosine function rule.

$\,\,\,\therefore\,\,\,\,\,\,$ $v \,=\, \sin{x}$

### Expand Definite integral by Integration by parts

Let us write the integration by parts formula in definite integral form.

$\displaystyle \int_{a}^{b}{u}\,dv$ $\,=\,$ $\big(uv\big)_{a}^{b}$ $-$ $\displaystyle \int_{a}^{b}{v}\,du$

The elements in the definite integration by parts formula are assumed as follows.

$(1).\,\,$ $u \,=\, x$

$(2).\,\,$ $v \,=\, \sin{x}$

$(3).\,\,$ $du \,=\, dx$

$(4).\,\,$ $dv \,=\, \cos{x}\,dx$

In this definite integral problem, the lower limit or bound $a \,=\, 0$ and the upper limit or bound $b \,=\, {\pi}/2$. Now, substitute them in the definite integration by parts rule.

$\implies$ $\displaystyle \int_{0}^{\Large \frac{\pi}{2}}{x\cos{x}}\,dx$ $\,=\,$ $\big(x\sin{x}\big)_{0}^{\Large \frac{\pi}{2}}$ $-$ $\displaystyle \int_{0}^{\Large \frac{\pi}{2}}{\sin{x}}\,dx$

### Evaluate the Definite integral of the function

Now, it is time to evaluate the definite integral of the product of the functions $x$ and $\cos{x}$ with respect to $x$ from $0$ to $\dfrac{\pi}{2}$.

$\,\,=\,\,\,$ $\big(x\sin{x}\big)_{0}^{\Large \frac{\pi}{2}}$ $-$ $\displaystyle \int_{0}^{\Large \frac{\pi}{2}}{\sin{x}}\,dx$

$\,\,=\,\,\,$ $\bigg(\dfrac{\pi}{2}\sin{\Big(\dfrac{\pi}{2}\Big)}-0\sin{0}\bigg)$ $-$ $\displaystyle \int_{0}^{\Large \frac{\pi}{2}}{\sin{x}}\,dx$

$\,\,=\,\,\,$ $\bigg(\dfrac{\pi}{2} \times \sin{\Big(\dfrac{\pi}{2}\Big)}-0 \times \sin{0}\bigg)$ $-$ $\displaystyle \int_{0}^{\Large \frac{\pi}{2}}{\sin{x}}\,dx$

According to the trigonometry, the sine of pi divided by two radian is equal to one, and the sine of angle zero radian is equal to zero.

$\,\,=\,\,\,$ $\Big(\dfrac{\pi}{2} \times 1-0 \times 0\Big)$ $-$ $\displaystyle \int_{0}^{\Large \frac{\pi}{2}}{\sin{x}}\,dx$

$\,\,=\,\,\,$ $\Big(\dfrac{\pi}{2}-0\Big)$ $-$ $\displaystyle \int_{0}^{\Large \frac{\pi}{2}}{\sin{x}}\,dx$

$\,\,=\,\,\,$ $\Big(\dfrac{\pi}{2}\Big)$ $-$ $\displaystyle \int_{0}^{\Large \frac{\pi}{2}}{\sin{x}}\,dx$

$\,\,=\,\,\,$ $\dfrac{\pi}{2}$ $-$ $\displaystyle \int_{0}^{\Large \frac{\pi}{2}}{\sin{x}}\,dx$

According to the integral rule of sine function, the definite integral of sine of angle $x$ with respect to $x$ can be calculated from lower bound $0$ to upper bound $\dfrac{\pi}{2}$.

$\,\,=\,\,\,$ $\dfrac{\pi}{2}$ $-$ $\big(-\cos{x}\big)_{0}^{\Large \frac{\pi}{2}}$

$\,\,=\,\,\,$ $\dfrac{\pi}{2}$ $+$ $\big(\cos{x}\big)_{0}^{\Large \frac{\pi}{2}}$

$\,\,=\,\,\,$ $\dfrac{\pi}{2}$ $+$ $\bigg(\cos{\Big(\dfrac{\pi}{2}\Big)}-\cos{(0)}\bigg)$

The cosine of pi by two radian is zero and the cosine of zero radian is one as per the trigonometry.

$\,\,=\,\,\,$ $\dfrac{\pi}{2}$ $+$ $(0-1)$

$\,\,=\,\,\,$ $\dfrac{\pi}{2}$ $+$ $(-1)$

$\,\,=\,\,\,$ $\dfrac{\pi}{2}-1$

$\,\,=\,\,\,$ $\dfrac{\pi-2}{2}$

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