/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 58 Evaluate \(\int_{0}^{\pi^{2} / 4... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 58

Evaluate \(\int_{0}^{\pi^{2} / 4} \sin \sqrt{x} d x\) using a substitution followed by integration by parts.

Short Answer

Expert verified
Based on the step-by-step solution provided, the definite integral \(\int_{0}^{\pi^{2} / 4} \sin \sqrt{x} dx\) is evaluated by using the substitution method followed by integration by parts. After simplifying, the final answer is found to be \(\boxed{\pi}\).

Step by step solution

01

Perform substitution

Let \(u = \sqrt{x}\), so \(x = u^2\). We also need to find the derivative of \(x\) with respect to \(u\). \(dx = 2u \ du\). Now let's find the new limits of the integral. When \(x = 0\), we get \(u = \sqrt{0} = 0\). When \(x = \frac{\pi^2}{4}\), we get \(u = \sqrt{\frac{\pi^2}{4}} = \frac{\pi}{2}\). Applying the substitution, our integral becomes: \(\int_{0}^{\pi/2} \sin u (2u \ du)\).
02

Apply integration by parts

To integrate by parts, we require two functions. Let \(v = \sin u \ du\) and \(dw = 2u \ du\). We now need to find \(dv\) and \(w\) by differentiating and integrating, respectively. By differentiating \(v\), we get: \(dv = \cos u \ du\). By integrating \(dw\), we get: \(w = u^2\). Now we can apply integration by parts using the formula: \(\int vd(w) = vw - \int wd(v)\). This gives us: \((u^2\sin u)\Big|_{0}^{\pi/2} - \int_{0}^{\pi/2} u^2 \cos u \ du\). Calculating \(u^2 \sin u\) from \(0\) to \(\pi/2\): \(\left(\frac{\pi^2}{4}\sin\frac{\pi}{2}\right) - (0) = \frac{\pi^2}{4}\).
03

Apply integration by parts again

To integrate \(u^2 \cos u \ du\), we apply integration by parts again. This time, let \(v = u^2 \ du\) and \(dw = \cos u \ du\). We need to find \(dv\) and \(w\). By differentiating \(v\), we get: \(dv = 2u \ du\). By integrating \(dw\), we get: \(w = \sin u\). Applying integration by parts formula again: \(\int u^2 \cos u \ du = u^2 \sin u - \int 2u \sin u \ du\).
04

Apply integration by parts once more

Now, we need to integrate \(2u \sin u \ du\). Let \(v = 2u \ du\) and \(dw = \sin u \ du\). We need to find \(dv\) and \(w\). By differentiating \(v\), we get: \(dv = 2 \ du\). By integrating \(dw\), we get: \(w = -\cos u\). Applying integration by parts formula once more: \(\int 2u \sin u \ du = -2u\cos u - \int{-2\cos u \ du}\). Now it's easy to integrate the last term: \(-2u\cos u + 2\sin u\Big|_{0}^{\pi/2} = -2 \frac{\pi}{2}\).
05

Combine all results

Now, we combine all the results from the previous steps: \(\frac{\pi^2}{4} - \int_{0}^{\pi/2} u^2 \cos u \ du = \frac{\pi^2}{4} - \left(u^2 \sin u - \int 2u \sin u \ du\right) \Big|_{0}^{\pi/2}\). We substitute the result from step 4: \(\frac{\pi^2}{4} - \left(\left(\frac{\pi}{2}\right)^2 \sin \frac{\pi}{2} - (-\pi)\right) = \frac{\pi^2}{4} - \left(\frac{\pi^2}{4} - \pi\right)\). After simplifying the expression, we get the final answer: \(\int_{0}^{\pi^{2} / 4} \sin \sqrt{x} d x = \boxed{\pi}\).

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