Math Doubts

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.

definite integration by parts question 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.

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

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

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

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

Math Questions

The math problems with solutions to learn how to solve a problem.

Learn solutions

Math Worksheets

The math worksheets with answers for your practice with examples.

Practice now

Math Videos

The math videos tutorials with visual graphics to learn every concept.

Watch now

Subscribe us

Get the latest math updates from the Math Doubts by subscribing us.

Learn more

Math Doubts

A free math education service for students to learn every math concept easily, for teachers to teach mathematics understandably and for mathematicians to share their maths researching projects.

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved