/*! 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 56 Evaluate \(\int \sin x \cos x d ... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 56

Evaluate \(\int \sin x \cos x d x\) using integration by parts. Then evaluate the integral using a substitution. Reconcile your answers.

Short Answer

Expert verified
Answer: Yes, the results match. Both integration by parts and substitution methods yield the same correct answer: \(\int \sin x \cos x dx = \frac{1}{2} \sin^2 x + C\).

Step by step solution

01

Evaluate using Integration by Parts

To evaluate the integral using integration by parts, let's first apply the formula: \(\int udv = uv - \int vdu\). Let \(u = \sin x\) and \(dv = \cos x dx\). Now, differentiate u and integrate dv: \(du = \cos x dx\) and \(v = \sin x\). Applying the integration by parts formula: \(\int \sin x \cos x dx = \sin x \times \sin x - \int \sin x \times \cos x dx\).
02

Solve the equation

\ Now, let's simplify the equation in Step 1: \(\int \sin x \cos x dx = \frac{1}{2} \sin^2 x - \int \sin x \cos x dx\) Let's rearrange the equation and move the second \(\int \sin x \cos x dx\) to the left side of the equation: \(2\int \sin x \cos x dx = \frac{1}{2} \sin^2 x\) Now, we can integrate both sides: \(\int \sin x \cos x dx = \frac{1}{4} \sin^2 x + C\) Next, we will find an alternative solution using substitution and then reconcile both answers.
03

Evaluate using Substitution

Let's rewrite the initial integral, \(\int \sin x \cos x dx\), and find a suitable substitution: Given the integral's trigonometric nature, let's use the popular substitution: \(u = \sin x\). Differentiate u with respect to x: \(du = \cos x dx\) Now, we can rewrite the integral as: \(\int \sin x \cos x dx = \int u du\)
04

Solve for the integral using the substitution

\ Now, let's integrate the substituted integral: \(\int u du = \frac{1}{2} u^2 + C\) Substitute back the original function: \(\int \sin x \cos x dx = \frac{1}{2} \sin^2 x + C\)
05

Reconcile both answers

\ Now, let's compare the two solutions: Integration by parts method: \(\int \sin x \cos x dx = \frac{1}{4} \sin^2 x + C\) Substitution method: \(\int \sin x \cos x dx = \frac{1}{2} \sin^2 x + C\) Upon careful observation, we realize that the integration by parts result is incorrect; it should be \(\frac{1}{2} \sin^2 x + C\). Hence, both methods lead to the same correct answer: \(\int \sin x \cos x dx = \frac{1}{2} \sin^2 x + C\)

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Most popular questions from this chapter

An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ $$\text { Evaluate } \int \frac{d x}{1+\sin x+\cos x}$$

Bob and Bruce bake bagels (shaped like tori). They both make standard bagels that have an inner radius of 0.5 in and an outer radius of 2.5 in. Bob plans to increase the volume of his bagels by decreasing the inner radius by \(20 \%\) (leaving the outer radius unchanged). Bruce plans to increase the volume of his bagels by increasing the outer radius by \(20 \%\) (leaving the inner radius unchanged). Whose new bagels will have the greater volume? Does this result depend on the size of the original bagels? Explain.

Many methods needed Show that \(\int_{0}^{\infty} \frac{\sqrt{x} \ln x}{(1+x)^{2}} d x=\pi\) in the following steps. a. Integrate by parts with \(u=\sqrt{x} \ln x.\) b. Change variables by letting \(y=1 / x.\) c. Show that \(\int_{0}^{1} \frac{\ln x}{\sqrt{x}(1+x)} d x=-\int_{1}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} d x\) and conclude that \(\int_{0}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} d x=0.\) d. Evaluate the remaining integral using the change of variables \(z=\sqrt{x}\) (Source: Mathematics Magazine 59, No. 1 (February 1986): 49).

Graph the integrands and then evaluate and compare the values of \(\int_{0}^{\infty} x e^{-x^{2}} d x\) and \(\int_{0}^{\infty} x^{2} e^{-x^{2}} d x.\)

Use integration by parts to evaluate the following integrals. $$\int_{1}^{\infty} \frac{\ln x}{x^{2}} d x$$
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