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Evaluate the definite integral that given,[tex] \displaystyle \int\limits^{\frac{1}{2}}_{0}\dfrac{1}{\sqrt{x^{2}+1}} \, \mathrm

Berikut ini adalah pertanyaan dari samuel312021058 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Evaluate the definite integral that given, \displaystyle \int\limits^{\frac{1}{2}}_{0}\dfrac{1}{\sqrt{x^{2}+1}} \, \mathrm dx !a. \varphi
b \ln(\varphi)
c. \dfrac{\varphi}{\pi}
d. \sqrt{\pi}
e. \infty ​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Answer:

B. \ln(\varphi)

Explanation of the Steps:

To evaluate the definite integral, \displaystyle \int\limits^{\frac{1}{2}}_{0}\dfrac{1}{\sqrt{x^{2}+1}}\,\mathrm dx. We will use the integration by trigonometric substitution technique for the radical expression, \sqrt{x^2+1}. Look at the table where I've shown you how to use the trigonometric substitution for each radical expression in the draft. Well, we see the radical expression is in the form of \sqrt{x^2+a^2}. It means we'll perform the trigonometric substitution by taking x= \tan(\theta). Here are the steps.

Supposed that, \begin{aligned} x &= \tan(\theta) \implies \mathrm dx &= \sec^2(\theta) \end{aligned}. This substitution is invertible over   0\le \theta \le \arctan\left(\dfrac{1}{2} \right) the inverse tangent function and will give new limits for both sides.

\begin{aligned} \displaystyle \int\limits^{\frac{1}{2}}_{0}\dfrac{1}{\sqrt{x^{2}+1}} \,\mathrm{d}x &= \int\limits_{0}^{\arctan\left( \frac{1}{2} \right)} \sqrt{\sec^2(\theta)} \,\mathrm d\theta \\ &= \int\limits_{0}^{\arctan\left( \frac{1}{2} \right)} \sec(\theta) \,\mathrm d\theta \implies \int\limits_{0}^{\arctan\left( \frac{1}{2} \right)}\frac{\sec^2(\theta)+\sec(\theta)\tan(\theta)}{\tan(\theta)+\sec(\theta)} \,\mathrm d\theta \end{aligned}

For \dfrac{\sec^2(\theta)+\sec(\theta)\tan(\theta)}{\tan(\theta)+\sec(\theta)}, let's do the algebraic substitution by taking \sec(\theta)+\tan(\theta) \implies u.

Let, \begin{aligned} u &= \tan(\theta)+\sec(\theta) \\ \frac{\mathrm{d}u}{\mathrm{d}\theta} &=\sec(\theta)\tan(\theta)+\sec^2(\theta) \end{aligned} \qquad \begin{aligned} \mathrm du &= \sec(\theta)\tan(\theta)+\sec^2(\theta) \,\mathrm{d}\theta \\ \mathrm{d}\theta &=\frac{\mathrm{d}u}{\sec(\theta)\tan(\theta)+\sec^2(\theta)} \end{aligned}

Thus,

\begin{aligned} \displaystyle \int\limits_0^{\arctan\left( \frac{1}{2} \right) } \frac{\sec^2(\theta)+\sec(\theta)\tan(\theta)}{\tan(\theta)+\sec(\theta)} \,\mathrm{d}\theta &=\int\limits_1^{\frac{1+\sqrt{5}}{2}} \frac{\mathrm du}{u} \\ &=\ln(u)\Bigg|^{\frac{1+\sqrt{5}}{2}}_1 \\ &=\ln\left( \frac{1+\sqrt{5}}{2} \right) \\ &=\ln(\varphi) \qquad \because\varphi=\frac{1+\sqrt5}{2} =1.\overline{61803398875} \end{aligned}

where \varphi =\text{Golden Ratio}.

Then, the correct answer is B. \ln(\varphi)

Answer:B. [tex]\ln(\varphi)[/tex]Explanation of the Steps:To evaluate the definite integral, [tex]\displaystyle \int\limits^{\frac{1}{2}}_{0}\dfrac{1}{\sqrt{x^{2}+1}}\,\mathrm dx[/tex]. We will use the integration by trigonometric substitution technique for the radical expression, [tex]\sqrt{x^2+1}[/tex]. Look at the table where I've shown you how to use the trigonometric substitution for each radical expression in the draft. Well, we see the radical expression is in the form of [tex]\sqrt{x^2+a^2}[/tex]. It means we'll perform the trigonometric substitution by taking [tex]x= \tan(\theta)[/tex]. Here are the steps.Supposed that, [tex]\begin{aligned} x &= \tan(\theta) \implies \mathrm dx &= \sec^2(\theta) \end{aligned}[/tex]. This substitution is invertible over   [tex]0\le \theta \le \arctan\left(\dfrac{1}{2} \right)[/tex] the inverse tangent function and will give new limits for both sides.[tex]\begin{aligned} \displaystyle \int\limits^{\frac{1}{2}}_{0}\dfrac{1}{\sqrt{x^{2}+1}} \,\mathrm{d}x &= \int\limits_{0}^{\arctan\left( \frac{1}{2} \right)} \sqrt{\sec^2(\theta)} \,\mathrm d\theta \\ &= \int\limits_{0}^{\arctan\left( \frac{1}{2} \right)} \sec(\theta) \,\mathrm d\theta \implies \int\limits_{0}^{\arctan\left( \frac{1}{2} \right)}\frac{\sec^2(\theta)+\sec(\theta)\tan(\theta)}{\tan(\theta)+\sec(\theta)} \,\mathrm d\theta \end{aligned}[/tex]For [tex]\dfrac{\sec^2(\theta)+\sec(\theta)\tan(\theta)}{\tan(\theta)+\sec(\theta)}[/tex], let's do the algebraic substitution by taking [tex]\sec(\theta)+\tan(\theta) \implies u[/tex]. Let, [tex]\begin{aligned} u &= \tan(\theta)+\sec(\theta) \\ \frac{\mathrm{d}u}{\mathrm{d}\theta} &=\sec(\theta)\tan(\theta)+\sec^2(\theta) \end{aligned} \qquad \begin{aligned} \mathrm du &= \sec(\theta)\tan(\theta)+\sec^2(\theta) \,\mathrm{d}\theta \\ \mathrm{d}\theta &=\frac{\mathrm{d}u}{\sec(\theta)\tan(\theta)+\sec^2(\theta)} \end{aligned}[/tex]Thus,[tex]\begin{aligned} \displaystyle \int\limits_0^{\arctan\left( \frac{1}{2} \right) } \frac{\sec^2(\theta)+\sec(\theta)\tan(\theta)}{\tan(\theta)+\sec(\theta)} \,\mathrm{d}\theta &=\int\limits_1^{\frac{1+\sqrt{5}}{2}} \frac{\mathrm du}{u} \\ &=\ln(u)\Bigg|^{\frac{1+\sqrt{5}}{2}}_1 \\ &=\ln\left( \frac{1+\sqrt{5}}{2} \right) \\ &=\ln(\varphi) \qquad \because\varphi=\frac{1+\sqrt5}{2} =1.\overline{61803398875} \end{aligned}[/tex]where [tex]\varphi =\text{Golden Ratio}[/tex].Then, the correct answer is B. [tex]\ln(\varphi)[/tex]

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Last Update: Tue, 23 May 23
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