Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Use a computer algebra system to evaluate the following indefinite integrals. Assume that a is a positive real number. $$\int\left(a^{2}-x^{2}\right)^{3 / 2} d x$$

Short Answer

Expert verified
Question: Find the indefinite integral of \((a^2-x^2)^{3/2}\)dx where a is a positive real number. Answer: The indefinite integral of \((a^2-x^2)^{3/2}\)dx is given by the expression: $$a^4\left(\frac{1}{4}\arcsin\left(\frac{x}{a}\right)+\frac{1}{2}\sin\left(2\arcsin\left(\frac{x}{a}\right)\right)+\frac{1}{8}\left(\arcsin\left(\frac{x}{a}\right)+\frac{1}{2}\sin\left(4\arcsin\left(\frac{x}{a}\right)\right)\right)\right) + C$$

Step by step solution

01

Choose a substitution

Let us choose the substitution \(x=a\sin(u)\), where u is a new variable. This substitution is chosen because it will simplify the square root located inside the expression.
02

Calculate the derivative of the substitution

Now we need to find the derivative of our substitution with respect to the new variable u, so we can perform a change of variables in the integral. The derivative is: $$\frac{dx}{du}=a\cos(u)$$
03

Rewrite the integral using the substitution

Let us rewrite the integral using our substitution \(x=a\sin(u)\) and its derivative \(dx=a\cos(u) du\). Replacing x in the original integral: $$\int\left(a^{2}-x^{2}\right)^{3 / 2} d x = \int\left(a^2-a^2\sin^2(u)\right)^{3/2}a\cos(u)du$$
04

Simplify the integrand

Next, we need to simplify the expression inside the integral: $$\int\left(a^2\left(1-\sin^2(u)\right)\right)^{3/2}a\cos(u)du$$ Recall that \(\cos^2(u) = 1-\sin^2(u)\). Thus, we can replace the expression inside the parentheses with \(\cos^2(u)\): $$\int a^3\cos^3(u)\cdot a\cos(u)du$$
05

Rewrite the integral as a single power of cosine

Rewrite the integral as a single power of the cosine function: $$a^4 \int \cos^4(u)du$$
06

Use the double angle identity for cosine

To proceed, we need to use the double angle identity for cosine, which states that: $$\cos^2(x)=\frac{1+\cos(2x)}{2}$$ Now square the identity for cosine and replace \(\cos^4(u)\) with its equivalent expression: $$a^4\int\left(\frac{1+\cos(2u)}{2}\right)^2 du$$
07

Expand and integrate

Expand the expression inside the integral and integrate each term separately: $$a^4\int\left(\frac{1}{4}+\frac{1}{2}\cos(2u)+\frac{1}{4}\cos^2(2u)\right) du$$ Now integrating each component, we get: $$\frac{1}{4}a^4(u+\frac{1}{2}\sin(2u))+\frac{1}{4}a^4\int\cos^2(2u)du$$
08

Apply the double angle identity again and integrate

Use the double angle identity again for the last term, replacing \(\cos^2(2u)\) with its equivalent expression and integrate: $$a^4\left(\frac{1}{4}u+\frac{1}{2}\sin(2u)+\frac{1}{8}\left(u+\frac{1}{2}\sin(4u)\right)\right) + C$$ where C is the integration constant.
09

Substitute back the original variable and simplify

Finally, we need to substitute back our original variable x using the substitution \(x=a\sin(u)\). To find the inverse substitution, we use: $$u=\arcsin\left(\frac{x}{a}\right)$$ Now, replace u in the expression: $$a^4\left(\frac{1}{4}\arcsin\left(\frac{x}{a}\right)+\frac{1}{2}\sin\left(2\arcsin\left(\frac{x}{a}\right)\right)+\frac{1}{8}\left(\arcsin\left(\frac{x}{a}\right)+\frac{1}{2}\sin\left(4\arcsin\left(\frac{x}{a}\right)\right)\right)\right) + C$$ This is the indefinite integral of the given expression.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use the indicated substitution to convert the given integral to an integral of a rational function. Evaluate the resulting integral. $$\int \frac{d x}{\sqrt[4]{x+2}+1} ; x+2=u^{4}$$

Decaying oscillations Let \(a>0\) and \(b\) be real numbers. Use integration to confirm the following identities. a. \(\int_{0}^{\infty} e^{-a x} \cos b x d x=\frac{a}{a^{2}+b^{2}}\) b. \(\int_{0}^{\infty} e^{-a x} \sin b x d x=\frac{b}{a^{2}+b^{2}}\)

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. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ $$\text { Evaluate } \int \frac{d \theta}{\cos \theta-\sin \theta}$$

By reduction formula 4 in Section 3 $$\int \sec ^{3} u d u=\frac{1}{2}(\sec u \tan u+\ln |\sec u+\tan u|)+C$$ Graph the following functions and find the area under the curve on the given interval. $$f(x)=\left(9-x^{2}\right)^{-2},\left[0, \frac{3}{2}\right]$$

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?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free