W02 Stepwise

Due date: Thursday 1/22, 11:59pm

Trig power products

01

01

Somewhat odd power product

Compute the integral:

$$\int {\sin }^{2}x\cdot {\cos }^{3}xdx$$

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Solution

Solutions - 0040-01

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

$${\sin }^{2}x\cdot {\cos }^{3}x\gg \gg {\sin }^{2}x(1-{\sin }^{2}x)\cos x$$

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

$$\begin{array}{cc}\int {\sin }^{2}x(1-{\sin }^{2}x)\cos xdx& \gg \gg \int u^2(1-u^2)du\\ & \\ & \gg \gg \frac{u^3}{3}-\frac{u^5}{5}+C\\ & \\ & \gg \gg \frac{{\sin }^{3}x}{3}-\frac{{\sin }^{5}x}{5}+C\end{array}$$Link to original

02

02

Tangent and secant both even

Compute the integral:

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

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Solution

Solutions - 0040-02

Notice ${\sec }^{2}xdx$. Therefore integrate with $u$-sub setting $u=\tan x$ and $du={\sec }^{2}xdx$:

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

03

03

All even power product

Compute the integral:

$$\int {\sin }^{4}x\cdot {\cos }^{2}xdx$$

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Solution

Solutions - 0040-03

(1) Notice all even powers. Use power-to-frequency conversion:

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

Plug in:

$$\int {\sin }^{4}x\cdot {\cos }^{2}xdx\gg \gg \frac{1}{8}\int {(1-\cos 2x)}^2(1+\cos 2x)dx$$

Simplify:

$$\gg \gg \frac{1}{8}\int 1-\cos 2x-{\cos }^2(2x)+{\cos }^3(2x)dx$$

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(2) Reduce power again for ${\cos }^2(2x)$:

$${\cos }^2(2x)\gg \gg \frac{1}{2}(1+\cos (4x))$$

(This is derived from the power-to-frequency formula by changing ‘$x$’ to ‘$2x$’ in that formula.)

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(3) On the last term, swap even bunch:

$${\cos }^3(2x)\gg \gg (1-{\sin }^2(2x))\cos 2x$$

Plug all in and obtain:

$$\gg \gg \frac{1}{8}\int 1-\cos 2x-\frac{1}{2}(1+\cos 4x)+(1-{\sin }^2(2x))\cos 2xdx$$

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(4) Integrate the first three terms:

$$\begin{array}{c}\frac{1}{8}\int 1-\cos 2x-\frac{1}{2}(1+\cos 4x)dx\gg \gg \\ \frac{1}{8}x-\frac{1}{16}\sin 2x-\frac{1}{16}x-\frac{1}{64}\sin 4x+C\end{array}$$

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(5) Integrate the last term with $u$-sub, setting $u=\sin 2x$ and $du=2\cos 2xdx$:

$$\begin{array}{cc}\frac{1}{8}\int (1-{\sin }^2(2x))\cos 2xdx& \gg \gg \frac{1}{16}\int (1-{\sin }^2(2x))2\cos 2xdx\\ & \\ & \gg \gg \frac{1}{16}\int (1-u^2)du\\ & \\ & \gg \gg \frac{u}{16}-\frac{u^3}{48}+C\\ & \\ & \gg \gg \frac{\sin 2x}{16}-\frac{{\sin }^3(2x)}{48}+C\end{array}$$

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(6) Combine in final result:

$$\begin{array}{c}\int {\sin }^{4}x\cdot {\cos }^{2}xdx\gg \gg \\ \frac{1}{16}x-\frac{1}{64}\sin 4x-\frac{1}{48}{\sin }^3(2x)+C\end{array}$$

Note: It is also possible to rewrite ${\sin }^3(2x)$ using trig identities. So, equally valid answers may look different than this.

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Trig substitution

04

01

Trig sub

Compute the definite integral:

$${\int }_0^{1/2}\frac{x^2}{\sqrt{1-x^2}}dx$$

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Solution

Solutions - 0050-01

(1) Substitute $x=\sin \theta$ and thus $dx=\cos \theta d\theta$. Adjust the bounds as follows:

$$\begin{array}{cc}x& =0\gg \gg \theta =0\\ x& =\frac{1}{2}\gg \gg \theta =\frac{\pi }{6}\end{array}$$

Rewrite the integral:

$$\begin{array}{cccc}& {\int }_0^{1/2}\frac{x^2}{\sqrt{1-x^2}}dx& \gg \gg {\int }_0^{\pi /6}\frac{{\sin }^2\theta \cos \theta }{\sqrt{1-{\sin }^2\theta }}d\theta & \\ & & & \\ & & \gg \gg {\int }_0^{\pi /6}{\sin }^2\theta d\theta & \text{(Note A)}\end{array}$$

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(2) Use power-to-frequency conversion:

$$\begin{array}{cc}{\int }_0^{\pi /6}{\sin }^2\theta d\theta & \gg \gg \frac{1}{2}{\int }_0^{\pi /6}1-\cos 2\theta d\theta \\ & \\ & \gg \gg \frac{1}{2}{(\theta -\frac{1}{2}\sin 2\theta )|}_0^{\pi /6}\\ & \\ & \gg \gg \frac{\pi }{12}-\frac{\sin (\frac{\pi }{3})}{4}-\frac{1}{2}(0-0)\\ & \\ & \gg \gg \frac{\pi }{12}-\frac{\sqrt{3}}{8}\end{array}$$

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Note A: Use $1-{\sin }^2\theta ={\cos }^2\theta$, then $\sqrt{{\cos }^2\theta }=|\cos \theta |$ and this equals $\cos \theta$ for $0\le \theta \le \pi /6$.

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05

02

Trig sub

Compute the integral:

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

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Solution

Solutions - 0050-02

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

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

Then:

$$\begin{array}{c}{x}^2+4\gg \gg 4{\tan }^2\theta +4\\ \\ \gg \gg 4({\tan }^2\theta +1)\gg \gg 4{\sec }^2\theta \end{array}$$

Plug in:

$$\int \frac{dx}{\sqrt{x^2+4}}\gg \gg \int \frac{2{\sec }^2\theta }{\sqrt{4{\sec }^2\theta }}d\theta \gg \gg \int \sec \theta d\theta$$

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

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(2) Perform integration.

Either recall from memory, or multiply above and below by $\sec \theta +\tan \theta$, and obtain:

$$\int \sec \theta d\theta \gg \gg \ln |\tan \theta +\sec \theta |+C$$

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(3) To convert to $x$ we need $\sec \theta$ given that $\tan \theta =x/2$.

Draw triangle expressing $\tan \theta ={\displaystyle \frac{x}{2}}$:

Therefore $\sec \theta ={\displaystyle \frac{\sqrt{x^2+4}}{2}}$. We already know $\tan \theta ={\displaystyle \frac{x}{2}}$. Thus:

$$\int \sec \theta d\theta \gg \gg \ln |\frac{x}{2}+\frac{\sqrt{x^2+4}}{2}|+C$$

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(4) Simplify with log rules:

$$\begin{array}{c}\ln |\frac{x}{2}+\frac{\sqrt{x^2+4}}{2}|+C\gg \gg \ln \frac{1}{2}|x+\sqrt{x^2+4}|+C\\ \\ \gg \gg \ln \frac{1}{2}+\ln |x+\sqrt{x^2+4}|+C\gg \gg \ln |x+\sqrt{x^2+4}|+C\end{array}$$Link to original