W02 - Examples

Power product - odd power

Power product - odd power

Compute the integral:

$$\int {\cos }^{2}x\cdot {\sin }^{5}xdx$$

Solution

(1) Swap over the even bunch:

Max even bunch leaving power-one is ${\sin }^{4}x$.

$${\sin }^{5}x\gg \gg \sin x({\sin }^{2}x{)}^2\gg \gg \sin x(1-{\cos }^{2}x{)}^2$$

Apply to ${\sin }^{5}x$ in the integrand:

$$\int {\cos }^{2}x\cdot {\sin }^{5}xdx\gg \gg \int {\cos }^{2}x\cdot \sin x(1-{\cos }^{2}x{)}^{2}dx$$

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(2) Perform $u$-substitution on the power-one integrand:

Set $u=\cos x$. Hence $du=\sin xdx$. Recognize this in the integrand and convert:

$$\begin{array}{rl}\int {\cos }^{2}x\cdot \sin x(1-{\cos }^{2}x{)}^{2}dx& \gg \gg \int {\cos }^{2}x\cdot \left(1{\cos }^{2}x{)}^2\right(\sin xdx)\\ & \\ & \gg \gg \int u^2\cdot (1-u^2{)}^{2}du\end{array}$$

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(3) Integrate using power rule:

$$\int u^2\cdot (1-u^2{)}^{2}du=\frac{1}{3}{u}^3-\frac{2}{5}{u}^5+\frac{1}{7}{u}^7+C$$

Insert definition $u=\cos x$:

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

Power product - tan and sec

Power product - tan and sec

Compute the integral:

$$\int {\tan }^{5}x\cdot {\sec }^{3}xdx$$

Solution

(1) Try $du={\sec }^{2}xdx$:

Factor $du$ out of the integrand:

$$\int {\tan }^{5}x\cdot {\sec }^{3}xdx\gg \gg \int {\tan }^{5}x\cdot \sec x\left({\sec }^{2}xdx\right)$$

We then must swap over remaining $\sec x$ into the $\tan x$ type.

Cannot do this because $\sec x$ has odd power. Need even to swap.

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(2) Try again: $du=\sec x\tan xdx$:

Factor $du$ out of the integrand:

$$\int {\tan }^{5}x\cdot {\sec }^{3}xdx\gg \gg \int {\tan }^{4}x\cdot {\sec }^{2}x\left(\sec x\tan xdx\right)$$

Swap remaining $\tan x$ into $\sec x$ type:

$$\begin{array}{c}\gg \gg \int ({\tan }^{2}x{)}^2\cdot {\sec }^{2}x(\sec x\tan xdx)\\ \\ \gg \gg \int ({\sec }^{2}x-1{)}^2\cdot {\sec }^{2}x(\sec x\tan xdx)\end{array}$$

Substitute $u=\sec x$ and $du=\sec x\tan xdx$:

$$\gg \gg \int (u^2-1{)}^2\cdot u^{2}du$$

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(3) Integrate in $u$ and convert back to $x$:

$$\begin{array}{c}\gg \gg \int u^6-2u^4+u^{2}du\\ \\ \gg \gg \frac{u^7}{7}-2\frac{u^5}{5}+\frac{u^3}{3}+C\\ \\ \gg \gg \frac{{\sec }^{7}x}{7}-2\frac{{\sec }^{5}x}{5}+\frac{{\sec }^{3}x}{3}+C\end{array}$$ Link to original

Trig sub in quadratic - completing the square

Trig sub in quadratic - completing the square

Compute the integral:

$$\int \frac{dx}{\sqrt{x^2-6x+11}}$$

Solution

(1) Complete the square to obtain Pythagorean form:

$$x^2-6x+{\left(\frac{-6}{2}\right)}^2\gg \gg x^2-6x+9=(x-3{)}^2$$

Add and subtract to get desired constant term:

$$\begin{array}{c}{x}^2-6x+11\gg \gg x^2-6x+9-9+11\\ \\ \gg \gg (x-3{)}^2+2\end{array}$$

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(2) Perform shift substitution:

Set $u=x-3$ as inside the square:

$$(x-3{)}^2+2=u^2+2$$

Infer $du=dx$. Plug into integrand:

$$\int \frac{dx}{\sqrt{x^2-6x+11}}\gg \gg \int \frac{du}{\sqrt{u^2+2}}$$

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(3) Trig sub with $\tan \theta$:

Use substitution $u=\sqrt{2}\tan \theta$. (From triangle or memorized tip.)

Infer $du=\sqrt{2}{\sec }^2\theta d\theta$. Plug in data:

$$\int \frac{du}{\sqrt{u^2+2}}\gg \gg \int \frac{{\sec }^2\theta }{\sec \theta }d\theta =\int \sec \theta d\theta$$

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(4) Integrate using ad hoc memorized formula:

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

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(5) Convert trig back to $x$:

First in terms of $u$, referring to the triangle:

$$\tan \theta =\frac{u}{\sqrt{2}},\sec \theta =\frac{\sqrt{u^2+2}}{\sqrt{2}}$$

Then in terms of $x$ using $u=x-3$. Plug everything in:

$$\ln |\tan \theta +\sec \theta |+C\gg \gg \ln \left|\frac{x-3}{\sqrt{2}}+\frac{\sqrt{(x-3{)}^2+2}}{\sqrt{2}}\right|+C$$

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(6) Simplify using log rules:

$$\ln \frac{f(x)}{a}\gg \gg \ln f(x)-\ln a$$

The common denominator $\frac{1}{\sqrt{2}}$ can be pulled outside as $-\ln \sqrt{2}$.

The new term $-\ln \sqrt{2}$ can be “absorbed into the constant” (redefine $C$).

So we write our final answer thus:

$$\ln \left|x-3+\sqrt{(x-3{)}^2+2}\right|+C$$Link to original