W02 Regular

Due date: Sunday 1/25, 11:59pm

Trig power products

01 $★$

04

All odd power product

Compute the integral:

$$\int {\cos }^{7}xdx$$

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Solution

Solutions - 0040-04

(1) Notice odd power on $\cos x$. Swap the even bunch:

$$\begin{array}{c}{\cos }^{7}x\gg \gg {({\cos }^{2}x)}^3\cos x\\ \\ \gg \gg {(1-{\sin }^{2}x)}^3\cos x\end{array}$$

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(2) Perform $u$-sub setting $u=\sin x$ and thus $du=\cos xdx$:

$$\begin{array}{c}\int {\cos }^{7}xdx\gg \gg \int {(1-{\sin }^{2}x)}^3\cos xdx\\ \\ \gg \gg \int {(1-u^2)}^{3}du\gg \gg \int 1-3u^2+3u^4-u^{6}du\\ \\ \gg \gg u-u^3+3\frac{u^5}{5}-\frac{u^7}{7}+C\end{array}$$

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(3) Convert back to $x$:

$$\gg \gg \sin x-{\sin }^{3}x+\frac{3}{5}{\sin }^{5}x-\frac{1}{7}{\sin }^{7}x+C$$Link to original

02 $★$

05

Tangent and secant mixed parity

Compute the integral:

$$\int {\tan }^{3}x{\sec }^{2}xdx$$

(a) Using $du={\sec }^{2}xdx$.

(b) Using $du=\sec x\tan xdx$.

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Solution

Solutions - 0040-05

(a) Select $u=\tan x$ and thus $du={\sec }^{2}xdx$:

$$\begin{array}{cc}\int {\tan }^{3}x{\sec }^{2}xdx& \gg \gg \int u^{3}du\\ & \\ & \gg \gg \frac{u^4}{4}+C\\ & \\ & \gg \gg \frac{{\tan }^{4}x}{4}+C\end{array}$$

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(b)

(1) Select $u=\sec x$ and thus $du=\sec x\tan xdx$:

$$\int {\tan }^{3}x{\sec }^{2}xdx\gg \gg \int {\tan }^{2}x\sec x(\sec x\tan x)dx$$

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(3) Swap even bunch using $1+{\tan }^{2}x={\sec }^{2}x$:

$$\gg \gg \int ({\sec }^{2}x-1)\sec x(\sec x\tan x)dx$$

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(4) Perform $u$-sub with $u=\sec x$ and integrate:

$$\begin{array}{cc}& \gg \gg \int (u^2-1)udu\\ & \\ & \gg \gg \frac{u^4}{4}-\frac{u^2}{2}+C\\ & \\ & \gg \gg \frac{{\sec }^{4}x}{4}-\frac{{\sec }^{2}x}{2}+C\end{array}$$Link to original

03 $★$

06

Power product with negative power

Compute the integral:

$$\int \sin 7x{\sec }^{5}7xdx$$

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Solution

Solutions - 0040-06

(1) Change variable by substituting $y=7x$ and $dy=7dx$:

$$\begin{array}{c}\int \sin 7x{\sec }^{5}7xdx\gg \gg \frac{1}{7}\int \sin 7x{\sec }^{5}7x7dx\\ \\ \gg \gg \frac{1}{7}\int \sin y{\sec }^{5}ydy\end{array}$$

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(2) Identify $\sin y\sec y=\tan y$:

$$\gg \gg \frac{1}{7}\int \tan y{\sec }^{4}ydy$$

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(3) Perform $u$-sub with $u=\sec y$ and thus $du=\sec y\tan ydy$:

$$\begin{array}{c}\gg \gg \frac{1}{7}\int {\sec }^{3}y(\sec y\tan y)dy\gg \gg \frac{1}{7}\int u^{3}du\\ \\ \gg \gg \frac{u^4}{28}+C\gg \gg \frac{{\sec }^{4}y}{28}+C\gg \gg \frac{{\sec }^4(7x)}{28}+C\end{array}$$Link to original

Trig substitution

04 $★$

03

Trig sub

Compute the integral:

$$\int \frac{dx}{x^3\sqrt{x^2-4}}$$

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Solution

Solutions - 0050-03

(1) Notice $x^2-4$ pattern, so we should make use of the identity ${\sec }^{2}x-1={\tan }^{2}x$.

Select $x=2\sec \theta$ and thus $dx=2\sec \theta \tan \theta d\theta$. Then:

$$x^2-4\gg \gg 4{\sec }^2\theta -4\gg \gg 4{\tan }^{2}x$$

Plug in and simplify:

$$\begin{array}{c}\int \frac{dx}{x^3\sqrt{x^2-4}}\gg \gg \int \frac{2\sec \theta \tan \theta }{8{\sec }^3\theta |2\tan \theta |}d\theta \\ \\ \gg \gg \frac{1}{8}\int {\cos }^2\theta d\theta \end{array}$$

(We must assume that $\tan \theta >0$ for the relevant values of $\theta$ here.)

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(2) Use power-to-frequency conversion:

$$\begin{array}{c}\gg \gg \frac{1}{8}\int \frac{1}{2}(1+\cos 2\theta )d\theta \gg \gg \frac{1}{16}(\theta +\frac{1}{2}\sin 2\theta )+C\end{array}$$

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(3) Convert back to terms of $x$:

First draw a triangle expressing $\sec \theta ={\displaystyle \frac{x}{2}}$:

Therefore:

$$\sin \theta =\frac{2}{x},\cos \theta =\frac{\sqrt{x^2-4}}{x}$$

For $\sin 2\theta$, use the double-angle identity:

$$\sin 2\theta =2\sin \theta \cos \theta \gg \gg 2\frac{\sqrt{x^2-4}}{x}\frac{2}{x}$$

Therefore:

$$\begin{array}{c}\frac{1}{16}(\theta +\frac{1}{2}\sin 2\theta )+C\\ \\ \gg \gg \frac{1}{16}({\sec }^{-1}\frac{x}{2}+\frac{2\sqrt{x^2-4}}{x^2})+C\end{array}$$Link to original

05 $★$

04

Trig sub

Compute the integral:

$$\int \frac{dx}{\sqrt{x^2+4x+13}}$$

Hint: complete the square and then substitute.

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Solution

Solutions - 0050-04

(1) Complete the square:

$$\begin{array}{c}{x}^2+4x+13\gg \gg x^2+4x+{\left(\frac{4}{2}\right)}^2-{\left(\frac{4}{2}\right)}^2+13\\ \\ \gg \gg {(x+2)}^2+9\end{array}$$

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(2) Substitute $x+2=3\tan \theta$ and thus $dx=3{\sec }^2\theta d\theta$:

$$\begin{array}{c}\gg \gg \int \frac{dx}{\sqrt{{(x+2)}^2+9}}\gg \gg \int \frac{3{\sec }^2\theta d\theta }{\sqrt{9{\tan }^2\theta +9}}\\ \\ \gg \gg \int \sec \theta d\theta \gg \gg \ln |\tan \theta +\sec \theta |+C\end{array}$$

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(3) Convert back to terms of $x$:

First draw a triangle expressing $\tan \theta ={\displaystyle \frac{x+2}{3}}$:

It follows that $\sec \theta ={\displaystyle \frac{\sqrt{{(x+2)}^2+9}}{3}}$. Then:

$$\begin{array}{ccc}& \ln |\tan \theta +\sec \theta |+C& \\ & & \\ & \gg \gg \ln |\frac{x+2}{3}+\frac{\sqrt{{(x+2)}^2+9}}{3}|+C& \\ & & \\ & \gg \gg \ln |x+2+\sqrt{{(x+2)}^2+9}|+C& \text{(Note A)}\end{array}$$

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Note A: Using log rules, the denominator $3$ can be brought out as $-\ln 3$ which can be “absorbed” into the constant $C$.

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06

05

Trig sub

Compute the integral:

$$\int \frac{x^2}{{(x^2+1)}^{3/2}}dx$$

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Solution

Solutions - 0050-05

(1) Notice $x^2+1$ pattern, so we should make use of the identity ${\tan }^2\theta +1={\sec }^2\theta$.

Select $x=\tan \theta$ and thus $dx={\sec }^2\theta d\theta$. Then:

$${(x^2+1)}^{3/2}\gg \gg {({\sec }^2\theta )}^{3/2}\gg \gg {\sec }^3\theta$$

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(2) Convert to $\theta$ and integrate:

$$\begin{array}{c}\int \frac{x^2}{{(x^2+1)}^{3/2}}dx\gg \gg \int \frac{{\tan }^2\theta {\sec }^2\theta }{{\sec }^3\theta }d\theta \\ \\ \gg \gg \int \frac{{\tan }^2\theta }{\sec \theta }d\theta \gg \gg \int \frac{{\sec }^2\theta -1}{\sec \theta }d\theta \\ \\ \gg \gg \int \sec \theta -\cos \theta d\theta \gg \gg \ln |\tan \theta +\sec \theta |-\sin \theta +C\end{array}$$

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(3) Convert back to terms of $x$:

Draw a triangle expressing $\tan \theta ={\displaystyle \frac{x}{1}}$:

Therefore $\sec \theta =\sqrt{x^2+1}$ and $\sin \theta ={\displaystyle \frac{x}{\sqrt{x^2+1}}}$. Then:

$$\begin{array}{c}\ln |\tan \theta +\sec \theta |-\sin \theta +C\\ \\ \gg \gg \ln |x+\sqrt{x^2+1}|-\frac{x}{\sqrt{x^2+1}}+C\end{array}$$Link to original

07

06

Double sub: $u$-sub then trig sub

Compute the definite integral:

$${\int }_0^{\pi /2}\frac{\cos x}{\sqrt{1+{\sin }^{2}x}}dx$$

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Solution

Solutions - 0050-06

(1) Perform $u$-sub setting $u=\sin x$ and thus $du=\cos xdx$. Adjust the bounds as follows:

$$\begin{array}{cc}x& =0\gg \gg u=0\\ x& =\frac{\pi }{2}\gg \gg u=1\end{array}$$

Therefore:

$${\int }_0^{\pi /2}\frac{\cos x}{\sqrt{1+{\sin }^{2}x}}dx\gg \gg {\int }_0^1\frac{1}{\sqrt{1+u^2}}du$$

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(2) Notice $1+u^2$ pattern, so we should make use of the identity $1+{\tan }^2\theta ={\sec }^2\theta$.

Select $u=\tan \theta$ and thus $du={\sec }^2\theta d\theta$. Adjust bounds:

$$\begin{array}{cc}u& =0\gg \gg \theta =0\\ u& =1\gg \gg \theta =\frac{\pi }{4}\end{array}$$

Therefore:

$${\int }_0^1\frac{du}{\sqrt{1+u^2}}\gg \gg {\int }_0^{\pi /4}\frac{{\sec }^2\theta d\theta }{\sec \theta }\gg \gg {\int }_0^{\pi /4}\sec \theta d\theta$$

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(3) Integrate from memory or multiplying above and below by $\sec \theta +\tan \theta$:

$$\begin{array}{c}\gg \gg \ln |\sec \theta +\tan \theta |{|}_0^{\pi /4}\\ \\ \gg \gg \ln |\sec \frac{\pi }{4}+\tan \frac{\pi }{4}|-\ln |\sec 0+\tan 0|\\ \\ \gg \gg \ln |1+\sqrt{2}|-\ln 1\gg \gg \ln |1+\sqrt{2}|\end{array}$$Link to original

08

07

Trig sub for electric charge

A charged wire lies on the $x$-axis running from $x_1$ to $x_2$. The electric field at the point $p=(0,h)$ is given by:

$$E={\int }_{x_1}^{x_2}\frac{k\lambda h}{{(x^2+h^2)}^{3/2}}dx$$

Find the numerical value of $E$ assuming $\lambda =6.0\times {10}^{-4}$ and $k=8.99\times {10}^9$ and $h=3$ and $(x_1,x_2)=(-\mathrm{15,15})$.

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Solution

Solutions - 0050-07

(1) Take out constants and insert given values:

$${\int }_{x_1}^{x_2}\frac{k\lambda h}{{(x^2+h^2)}^{\frac{3}{2}}}dx\gg \gg k\lambda h{\int }_{-15}^{15}\frac{dx}{{(x^2+9)}^{\frac{3}{2}}}$$

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(2) Notice $x^2+9$ pattern, so we should make use of the identity ${\tan }^2\theta +1={\sec }^2\theta$.

Select $x=3\tan \theta$ and thus $dx=3{\sec }^2\theta d\theta$. Then:

$${(x^2+9)}^{3/2}\gg \gg {(9{\sec }^2\theta )}^{3/2}\gg \gg 27{\sec }^3\theta$$

Adjust bounds:

$$\begin{array}{cc}x& =-15\gg \gg \theta ={\tan }^{-1}(-5)\\ x& =15\gg \gg \theta ={\tan }^{-1}(5)\end{array}$$

Then:

$$\begin{array}{c}\gg \gg k\lambda h{\int }_{{\tan }^{-1}(-5)}^{{\tan }^{-1}(5)}\frac{3{\sec }^2\theta }{27{\sec }^3\theta }d\theta \\ \\ \gg \gg \frac{k\lambda h}{9}{\int }_{{\tan }^{-1}(-5)}^{{\tan }^{-1}(5)}\cos \theta d\theta \end{array}$$

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(3) Integrate:

$$\begin{array}{c}\gg \gg {\frac{k\lambda h}{9}\sin \theta |}_{{\tan }^{-1}(-5)}^{{\tan }^{-1}(5)}\\ \\ \gg \gg \frac{k\lambda h}{9}\left(\sin ({\tan }^{-1}(5))-\sin ({\tan }^{-1}(-5))\right)\\ \\ \gg \gg \frac{k\lambda h}{9}\left(\sin ({\tan }^{-1}(5))+\sin ({\tan }^{-1}(5))\right)\\ \\ \gg \gg \frac{(6\times {10}^{-4})(8.99\times {10}^9)}{3}2\sin ({\tan }^{-1}(5))\end{array}$$

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(4) Compute $\sin ({\tan }^{-1}(5))$:

Draw a triangle expressing $\tan \theta ={\displaystyle \frac{5}{1}}$:

Therefore $\sin ({\tan }^{-1}(5)={\displaystyle \frac{5}{\sqrt{26}}}$. Then:

$$\gg \gg \frac{(6\times {10}^{-3})(8.99\times {10}^{10})}{3\sqrt{26}}$$Link to original