/*! 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 7 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 7

Evaluate the following integrals. $$\int x \cos x d x$$

Short Answer

Expert verified
Question: Find the integral of the function \(f(x) = x \cos x\). Answer: The integral of \(f(x) = x \cos x\) with respect to \(x\) is given by \(\int x \cos x \, dx = x \sin x + \cos x + C\), where \(C\) is the constant of integration.

Step by step solution

01

Identify u and dv

We have chosen: $$u = x$$ and $$dv = \cos x \, dx$$
02

Differentiate u and integrate dv

We compute the derivative of \(u\) with respect to \(x\): $$du = \frac{du}{dx} dx = 1 \cdot dx = dx$$ And we integrate \(dv\): $$v = \int \cos x \, dx = \sin x + C_1$$ Here, \(C_1\) is a constant of integration which will cancel out during integration by parts.
03

Apply the integration by parts formula

We substitute the values of \(u\), \(v\), \(du\), and \(dv\) into the integration by parts formula: $$\int x \cos x \, dx = uv - \int v \, du = x(\sin x) - \int \sin x \, dx$$
04

Integrate sin x

Now we need to integrate \(\sin x\): $$\int \sin x \, dx = -\cos x + C_2$$ Where \(C_2\) is another constant of integration.
05

Substitute back into the main formula

Finally, we substitute the integration result into the main formula: $$\int x \cos x \, dx = x(\sin x) - (-\cos x + C_2)$$
06

Simplify and write the final answer

We simplify the expression and include a constant of integration \(C\): $$\int x \cos x \, dx = x \sin x + \cos x + C$$ Thus, the integral of \(x \cos x\) with respect to \(x\) is: $$\int x \cos x \, dx = x \sin x + \cos x + C$$

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Most popular questions from this chapter

Use the following three identities to evaluate the given integrals. $$\begin{aligned}&\sin m x \sin n x=\frac{1}{2}[\cos ((m-n) x)-\cos ((m+n) x)]\\\&\sin m x \cos n x=\frac{1}{2}[\sin ((m-n) x)+\sin ((m+n) x)]\\\&\cos m x \cos n x=\frac{1}{2}[\cos ((m-n) x)+\cos ((m+n) x)]\end{aligned}$$ $$\int \sin 5 x \sin 7 x d x$$

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).

A long, straight wire of length \(2 L\) on the \(y\) -axis carries a current \(I\). According to the Biot-Savart Law, the magnitude of the magnetic field due to the current at a point \((a, 0)\) is given by $$B(a)=\frac{\mu_{0} I}{4 \pi} \int_{-L}^{L} \frac{\sin \theta}{r^{2}} d y$$ where \(\mu_{0}\) is a physical constant, \(a>0,\) and \(\theta, r,\) and \(y\) are related as shown in the figure. a. Show that the magnitude of the magnetic field at \((a, 0)\) is $$B(a)=\frac{\mu_{0} I L}{2 \pi a \sqrt{a^{2}+L^{2}}}$$ b. What is the magnitude of the magnetic field at \((a, 0)\) due to an infinitely long wire \((L \rightarrow \infty) ?\)

Use integration by parts to evaluate the following integrals. $$\int_{0}^{1} x \ln x d x$$

Imagine that today you deposit \(\$ B\) in a savings account that earns interest at a rate of \(p \%\) per year compounded continuously. The goal is to draw an income of \(\$ I\) per year from the account forever. The amount of money that must be deposited is \(B=I \int_{0}^{\infty} e^{-n t} d t,\) where \(r=p / 100 .\) Suppose you find an account that earns \(12 \%\) interest annully and you wish to have an income from the account of \(\$ 5000\) per year. How much must you deposit today?
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