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 $\Big[0,\,\dfrac{\pi}{2}\Big]$.

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.

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

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.

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

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.

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

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

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

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

Latest Math Topics

Aug 31, 2024

Aug 07, 2024

Jul 24, 2024

Dec 13, 2023

Latest Math Problems

Oct 22, 2024

Oct 17, 2024

Sep 04, 2024

Jan 30, 2024

Oct 15, 2023

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved