The only difference is whether the integrand is positive or negative. &=\frac{41}{6} \end{align*}\]. In this topic, the student will learn the Integration concepts as well as some integration formula with examples… It is important to note that these formulas are presented in terms of indefinite integrals. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Follow the procedures from Examples \(\PageIndex{2}\) and \(\PageIndex{3}\). Recall, that trigonometric functions are not one-to-one unless the domains are restricted. Example \( \PageIndex{2}\): Finding an Antiderivative Involving an Inverse Trigonometric Function using substitution, \[ ∫\dfrac{dx}{\sqrt{4−9x^2}}.\nonumber\], Substitute \( u=3x\). Suppose a car is moving due north (the positive direction) at 40 mph between 2 p.m. and 4 p.m., then the car moves south at 30 mph between 4 p.m. and 5 p.m. We can graph this motion as shown in Figure \(\PageIndex{1}\). It says that when a quantity changes, the new value equals the initial value plus the integral of the rate of change of that quantity. \end{align*}\]. &=\left(5t−\frac{3t^2}{2}\right)\bigg|^{5/3}_0+\left(\frac{3t^2}{2}−5t\right)\bigg|^3_{5/3} \\[4pt] The net change theorem states that when a quantity changes, the final value equals the initial value plus the integral of the rate of change. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. To find net displacement, integrate the velocity function over the interval. We have worked with these functions before. &=\left[\frac{(2)^9}{3}−2(2)\right]−\left[\frac{(−2)^9}{3}−2(−2)\right] \\[4pt] Net change accounts for negative quantities automatically without having to write more than one integral. We have, \[ \begin{align*} ∫^2_0\left(5−t^3\right)\,dt &=\left(5t−\frac{t^4}{4}\right)∣^2_0 \\[4pt] &=\left[5(2)−\frac{(2)^4}{4}\right]−0 \\[4pt] &=10−\frac{16}{4} \\[4pt] &=6. The formulas developed there give rise directly to integration formulas involving inverse trigonometric functions. A definite integral is either a number (when the limits of integration are constants) or a single function (when one or both of the limits of integration are variables). Example \(\PageIndex{6}\): Integrating an Even Function. \end{align*}\], Find the definite integral of \( f(x)=x^2−3x\) over the interval \([1,3].\). In many integrals that result in inverse trigonometric functions in the antiderivative, we may need to use substitution to see how to use the integration formulas provided above. Recalling that Andrew’s iceboat travels at twice the wind speed, and assuming he moves in a straight line away from his starting point, how far is Andrew from his starting point after \(1\) hour? Follow the process from Example \(\PageIndex{1}\) to solve the problem. INTEGRATION OF TRIGONOMETRIC INTEGRALS . Follow the procedures from Example \(\PageIndex{6}\) to solve the problem. The area under an even function over a symmetric interval can be calculated by doubling the area over the positive \(x\)-axis. We saw in Functions and Graphs that an even function is a function in which \(f(−x)=f(x)\) for all \(x\) in the domain—that is, the graph of the curve is unchanged when \(x\) is replaced with \(−x\). &=\big[20t^2+10t\big]\bigg|^{1/2}_0+\big[30t\big]\bigg|^1_{1/2} \\[4pt] We can see the symmetry about the origin by the positive area above the \(x\)-axis over \([−π,0]\), and the negative area below the \(x\)-axis over \([0,π].\) we have, \[ \begin{align*} ∫^π_{−π}−5\sin x \,dx &=−5(−\cos x)\bigg|^π_{−π} \\[4pt] &=5\cos x\,\bigg|^π_{−π} \\[4pt] &=[5\cos π]−[5\cos(−π)] \\[4pt] &=−5−(−5)=0. Find the indefinite integral using an inverse trigonometric function and substitution for \(\displaystyle ∫\dfrac{dx}{\sqrt{9−x^2}}\). Graph (a) shows the region below the curve and above the \(x\)-axis. Download for free at http://cnx.org. The total distance traveled is given by, \[ ∫^5_2|v(t)|\,dt=∫^4_240\,dt+∫^5_430\,dt=80+30=110. For continuous even functions such that \(f(−x)=f(x),\), For continuous odd functions such that \(f(−x)=−f(x),\), Example \(\PageIndex{6}\): Integrating an Even Function. Under these conditions, how far from his starting point is Andrew after 1 hour? Let F(x) be any If the motor on a motorboat is started at \(t=0\) and the boat consumes gasoline at the rate of \(5−t^3\) gal/hr, how much gasoline is used in the first \(2\) hours? Since the function is negative over the interval \(\left[0,\frac{5}{3}\right]\), we have \(\big|v(t)\big|=−v(t)\) over that interval. Theorem Let f(x) be a continuous function on the interval [a,b]. Substitution is often required to put the integrand in the correct form. In this section we focus on integrals that result in inverse trigonometric functions. Let us begin this last section of the chapter with the three formulas. Legal. Evaluate the definite integral \(\displaystyle ∫^2_0\dfrac{dx}{4+x^2}\). Don’t forget that Andrew’s iceboat moves twice as fast as the wind. To illustrate, let’s apply the net change theorem to a velocity function in which the result is displacement. The following integration formulas yield inverse trigonometric functions: \[ \begin{align} ∫\dfrac{du}{\sqrt{a^2−u^2}} =\sin^{−1}\left(\dfrac{u}{a}\right)+C \\ ∫\dfrac{du}{a^2+u^2} =\dfrac{1}{a}\tan^{−1}\left(\dfrac{u}{a}\right)+C \\ ∫\dfrac{du}{u\sqrt{u^2−a^2}} =\dfrac{1}{a}\sec^{−1}\left(\dfrac{|u|}{a}\right)+C \end{align}\], Let \( y=\sin^{−1}\frac{x}{a}\). Also, we previously developed formulas for derivatives of inverse trigonometric functions. Area and perimeter. Recall the integration formulas given in the section on Antiderivatives and the properties of definite integrals. To close this section, we examine one more formula: the integral resulting in the inverse tangent function. The symmetry appears in the graphs in Figure \(\PageIndex{4}\). Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors.

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