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a) Find the indefinite integral using the substitution x = 7 \sec(?). (Use C for the constant of...

Question:

a) Find the indefinite integral using the substitution {eq}x = 7 \sec(?). {/eq} (Use C for the constant of integration.)

{eq}\int x^{3}\sqrt{x^{2}-49}dx {/eq}

and substitution {eq}x = 9 \tan(?) {/eq} for:

b) Find the volume of the torus generated by revolving the region bounded by the graph of the circle about the y-axis.

{eq}(x - 17)^{2} + y^{2} = 1 {/eq}

c) Find the indefinite integral. (Use C for the constant of integration.)

{eq}\int \sqrt{16-4^{2]}dx {/eq}

Indefinite Integral:


Integration is reverse of differentiation.

Here, we will integrate the function by doing trigonometric substitution.

We will add constant C at the end of evaluation.

Answer and Explanation:


Given integral is

a)

{eq}\displaystyle I=\int \:x^3\sqrt{x^2-49}dx\\ \displaystyle Substitute,\:x=7sec\left(u\right)\:==>dx=7sec\left(u\right)tan\left(u\right)du\\ \displaystyle I=\int \:7^3sec^3\left(u\right)\sqrt{7^2sec^2\left(u\right)-7^2}\:7sec\left(u\right)tan\left(u\right)du\:\\ \displaystyle I=\int \:\:7^5sec^4\left(u\right)\sqrt{tan^2\left(u\right)}\:tan\left(u\right)du\:\\ \displaystyle I=7^5\int \:sec^4\left(u\right)\:tan^2\left(u\right)du\:\\ \displaystyle I=7^5\int \:sec^2\left(u\right)\:tan^2\left(u\right)sec^2\left(u\right)du\:\\ \displaystyle I=7^5\int \:\left(1+tan^2\left(u\right)\right)\:tan^2\left(u\right)sec^2\left(u\right)du\:\\ \displaystyle I=7^5\int \:\left(tan^2\left(u\right)+tan^4\left(u\right)\right)\:sec^2\left(u\right)du\:\\ \displaystyle Substitute,\:tan\left(u\right)=t==>sec^2\left(u\right)du=dt\\ \displaystyle I=7^5\int \:\left(t^2+t^4\right)dt\:\\ \displaystyle I=7^5\left(\frac{t^3}{3}+\frac{t^5}{5}\right)+C\:\\ \displaystyle Resubstitute,\:t=tan\left(u\right)=\frac{\sqrt{x^2-49}}{7}\\ \displaystyle I=7^5\left(\frac{\left(\sqrt{x^2-49}\right)^3}{3\times 7^3}+\frac{\left(\sqrt{x^2-49}\right)^5}{5\times \:7^5}\right)+C\\ \displaystyle I=\frac{49}{3}\left(x^2-49\right)^{\frac{3}{2}}+\frac{1}{5}\left(x^2-49\right)^{\frac{5}{2}}+C\:\:\: {/eq}


b)

{eq}\displaystyle \mathrm{The\:volume\:for\:solid\:of\:revolution\:for\:a\:curve\:is\:the\:volume\:of\:object\:determined\:by\:a\:curve}\:f\left(x\right)\:\\ \mathrm{rotated\:around}\:\mathrm{the}\:x\mathrm{-axis}\:\mathrm{on\:an\:interval}\:\left[a,\:b\right]\:\mathrm{given\:by}\\ \displaystyle V=\int _a^b\pi \left(f\left(x\right)\right)^2dx\\ {/eq}

Given equation is

{eq}\displaystyle \left(x-17\right)^2+y^2=1\\ \displaystyle y^2=-x^2+34x-288\\ \displaystyle \mathrm{To\:find\:the\:intersection\:points\:solve}\:f\left(x\right)^2=0\\ \displaystyle \therefore\:-x^2+34x-288=0\\ \displaystyle x=16,\:x=18\\ \displaystyle \therefore\:V=\int _{16}^{18}\pi \left(-x^2+34x-288\right)dx\\ \displaystyle V=-1620\pi -\left(-\pi \frac{4864}{3}\right)\\ \displaystyle V=\frac{4\pi }{3} {/eq}


c)

{eq}\displaystyle I=\int \:\sqrt{16-x^2}dx\:\\ \displaystyle Substitute,\:x=4sin\left(u\right)==>dx=4cos\left(u\right)du\\ \displaystyle I=\int \:\sqrt{16-16sin^2\left(u\right)}4cos\left(u\right)du\:\\ \displaystyle I=16\int \:cos^2\left(u\right)du\:\\ \displaystyle I=16\int \:\frac{1+cos\left(2u\right)}{2}du\\ \displaystyle I=8\left[u+\frac{sin\left(2u\right)}{2}\right]+C\\ \displaystyle Resubstitute,\:sin\left(u\right)=\frac{x}{4}\\ \displaystyle I=8\left[sin^{-1}\left(\frac{x}{4}\right)+\frac{x\sqrt{16-x^2}}{16}\right]+C\: {/eq}


Learn more about this topic:

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Indefinite Integrals as Anti Derivatives

from Math 104: Calculus

Chapter 12 / Lesson 11
8.1K

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