5.5 Substitution

Solution: The first step is to choose an expression for $u$. We choose $u=3x^2+4$ because then $du=6xdx$, and we already have $du$ in the integrand. Write the integral in terms of $u$:

$$\int 6x{(3x^2+4)}^4𝑑x=\int u^4𝑑u.$$

Remember that $du$ is the derivative of the expression chosen for $u$, regardless of what is inside the integrand. Now we can evaluate the integral with respect to $u$:

	${\displaystyle \int u^4𝑑u}$	$={\displaystyle \frac{u^5}{5}}+C$
		$={\displaystyle \frac{{(3x^2+4)}^5}{5}}+C.$

Analysis: We can check our answer by taking the derivative of the last result and seeing if we get the integrand. Let $y={\displaystyle \frac{1}{5}}{(3x^2+4)}^5+C$. Then

	${\displaystyle \frac{d}{dx}}y$	$={\displaystyle \frac{d}{dx}}\left({\displaystyle \frac{1}{5}}{(3x^2+4)}^5+C\right)$
	${\displaystyle \frac{dy}{dx}}$	$={\displaystyle \frac{1}{5}}5{(3x^2+4)}^{4}6x+0$
		$=6x{(3x^2+4)}^4.$

This is exactly the expression we started with inside the integrand.

Solution: Rewrite the integral as ${\displaystyle \int z{(z^2-5)}^{1/2}𝑑z}$. Suppose we try $u=z^2-5$ and $du=2zdz$. Now we have a problem because $du=2zdz$ and the original expression has only $zdz$. We have to alter our expression for $du$ or the integral in $u$ will be twice as large as it should be. If we multiply both sides of the $du$ equation by ${\displaystyle \frac{1}{2}}$ we can solve this problem. Thus,

	$u$	$=z^2-5$
	$du$	$=2zdz$
	${\displaystyle \frac{1}{2}}du$	$={\displaystyle \frac{1}{2}}\left(2z\right)dz=zdz.$

Write the integral in terms of $u$, but pull the ${\displaystyle \frac{1}{2}}$ outside the integration symbol:

$$\int z{(z^2-5)}^{1/2}𝑑z=\frac{1}{2}\int u^{1/2}𝑑u.$$

Integrate the expression in $u$:

	${\displaystyle \frac{1}{2}}{\displaystyle \int u^{1/2}𝑑u}$	$=\left({\displaystyle \frac{1}{2}}\right){\displaystyle \frac{u^{3/2}}{\frac{3}{2}}}+C$
		$=\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{2}{3}}\right)u^{3/2}+C$
		$={\displaystyle \frac{1}{3}}{u}^{3/2}+C$
		$={\displaystyle \frac{1}{3}}{(z^2-5)}^{3/2}+C.$

Solution: Use substitution, setting $u=-x$, and then $du=-1dx$. Multiply the $du$ equation by $-$1, so you now have $-du=dx$. Then,

	${\displaystyle \int e^{-x}𝑑x}$	$=-{\displaystyle \int e^u𝑑u}$
		$=-e^u+C$
		$=-e^{-x}+C.$

Solution: We know the derivative of $\cos t$ is $-\sin t$, so we set $u=\cos t$. Then $du=-\sin tdt$. Substituting into the integral, we have

$$\int \frac{\sin t}{{\cos }^{3}t}𝑑t=-\int \frac{du}{u^3}.$$

Evaluating the integral, we get

	$-{\displaystyle \int \frac{du}{u^3}}$	$=-{\displaystyle \int u^{-3}𝑑u}$
		$=-\left(-{\displaystyle \frac{1}{2}}\right)u^{-2}+C.$

Putting the answer back in terms of $t$, we get

	${\displaystyle \int \frac{\sin t}{{\cos }^{3}t}𝑑t}$	$={\displaystyle \frac{1}{2u^2}}+C$
		$={\displaystyle \frac{1}{2{\cos }^{2}t}}+C.$

Solution: Let $u=1+2x^3$, so $du=6x^{2}dx$. Since the original function includes one factor of $x$² and $du=6x^{2}dx$, multiply both sides of the $du$ equation by $1/6$. Then,

	$du$	$=6x^{2}dx$
	${\displaystyle \frac{1}{6}}du$	$=x^{2}dx.$

To adjust the limits of integration, note that when $x=0$, then $u=1+2(0)=1$, and when $x=1$ then $u=1+2(1)=3$. So

$${\int }_0^1{x}^2{(1+2x^3)}^5𝑑x=\frac{1}{6}{\int }_1^3{u}^5𝑑u.$$

Evaluating this expression, we get

	${\displaystyle \frac{1}{6}}{\displaystyle {\int }_1^3}{u}^5𝑑u$	$={\left({\displaystyle \frac{1}{6}}\right)\left({\displaystyle \frac{u^6}{6}}\right)|}_1^3$
		$={\displaystyle \frac{1}{36}}\left[{(3)}^6-{(1)}^6\right]$
		$={\displaystyle \frac{182}{9}}.$

Solution: Let $u=4x^2+3$. Then, $du=8xdx$. To adjust the limits of integration, we note that when $x=0$ then $u=3$, and when $x=1$ then $u=7$. So our substitution gives

	${\displaystyle {\int }_0^1}x{e}^{4x^2+3}𝑑x$	$={\displaystyle \frac{1}{8}}{\displaystyle {\int }_3^7}{e}^u𝑑u$
		$={{\displaystyle \frac{1}{8}}{e}^u|}_3^7$
		$={\displaystyle \frac{e^7-e^3}{8}}$
		$\approx 134.568.$

Solution: First rewrite the problem using a rational exponent:

$$\int e^x\sqrt{1+e^x}𝑑x=\int e^x{(1+e^x)}^{1/2}𝑑x.$$

Using substitution, choose $u=1+e^x.u=1+e^x$. Then, $du=e^{x}dx$. We have (Figure 5.40)

$$\int e^x{(1+e^x)}^{1/2}𝑑x=\int u^{1/2}𝑑u.$$

Then

$$\int u^{1/2}𝑑u=\frac{u^{3/2}}{3/2}+C=\frac{2}{3}{u}^{3/2}+C=\frac{2}{3}{(1+e^x)}^{3/2}+C.$$

Solution: Here we choose to let $u$ equal the expression in the exponent on $e$. Let $u=2x^3$ and $du=6x^{2}dx.$. Again, $du$ is off by a constant multiplier; the original function contains a factor of 3$x$², not 6$x$². Multiply both sides of the equation by ${\displaystyle \frac{1}{2}}$ so that the integrand in $u$ equals the integrand in $x$. Thus,

$$\int 3x^2{e}^{2x^3}𝑑x=\frac{1}{2}\int e^u𝑑u.$$

Integrate the expression in $u$ and then substitute the original expression in $x$ back into the $u$ integral:

$$\frac{1}{2}\int e^u𝑑u=\frac{1}{2}{e}^u+C=\frac{1}{2}{e}^{2x^3}+C.$$

Solution: To find the price–demand equation, integrate the marginal price–demand function. First find the antiderivative, then look at the particulars. Thus,

	$p(x)$	$={\displaystyle \int }-0.015e^{-0.01x}dx$
		$=-0.015{\displaystyle \int e^{-0.01x}𝑑x}.$

Using substitution, let $u=-0.01x$ and $du=-0.01dx$. Then, divide both sides of the $du$ equation by $-$0.01. This gives

	${\displaystyle \frac{-0.015}{-0.01}}{\displaystyle \int e^u𝑑u}$	$=1.5{\displaystyle \int e^u𝑑u}$
		$=1.5e^u+C$
		$=1.5e^{-0.01x}+C.$

The next step is to solve for $C$. We know that when the price is $2.35 per tube, the demand is 50 tubes per week. This means

	$p(50)$	$=1.5e^{-0.01(50)}+C$
		$=2.35.$

Now, just solve for $C$:

	$C$	$=2.35-1.5e^{-0.5}$
		$=2.35-0.91$
		$=1.44.$

Thus,

$$p(x)=1.5e^{-0.01x}+1.44.$$

If the supermarket sells 100 tubes of toothpaste per week, the price would be

$$p(100)=1.5e^{-0.01(100)}+1.44=1.5e^{-1}+1.44\approx 1.99.$$

The supermarket should charge $1.99 per tube if it is selling 100 tubes per week.

Solution: Again, substitution is the method to use. Let $u=1-x$, so $du=-1dx$ or $-du=dx$. Then ${\displaystyle \int e^{1-x}𝑑x}=-{\displaystyle \int e^u𝑑u}$. Next, change the limits of integration. Using the equation $u=1-x$, we have

	$u=1-(1)=0$
	$u=1-(2)=-1.$

The integral then becomes

	${\displaystyle {\int }_1^2}{e}^{1-x}𝑑x$	$=-{\displaystyle {\int }_0^{-1}}{e}^u𝑑u$
		$={\displaystyle {\int }_{-1}^0}{e}^u𝑑u$
		$={e^u|}_{-1}^0$
		$=e^0-(e^{-1})$
		$=-e^{-1}+1.$

See Figure 5.41.

Solution: We have

$$Q(t)=\int 3^t𝑑t=\frac{3^t}{\ln 3}+C.$$

Then, at $t=0$ we have $Q(0)=10={\displaystyle \frac{1}{\ln 3}}+C$, so $C\approx 9.090$ and we get

$$Q(t)=\frac{3^t}{\ln 3}+9.090.$$

At time $t=2$, we have

$$Q(2)=\frac{3^2}{\ln 3}+9.090$$

$$=17.282.$$

After 2 hours, there are 17,282 bacteria in the dish.

Solution: Let $G(t)$ represent the number of flies in the population at time $t$. Applying the net change theorem, we have

	$G(10)$	$=G(0)+{\displaystyle {\int }_0^{10}}2e^{0.02t}𝑑t$
		$=100+{\left[{\displaystyle \frac{2}{0.02}}{e}^{0.02t}\right]|}_0^{10}$
		$=100+{[100e^{0.02t}]|}_0^{10}$
		$=100+100e^{0.2}-100$
		$\approx 122.$

There are 122 flies in the population after 10 days.

Solution: This problem requires some rewriting to simplify applying the properties. First, rewrite the exponent on $e$ as a power of $x$, then bring the $x$² in the denominator up to the numerator using a negative exponent. We have

$${\int }_1^2\frac{e^{1/x}}{x^2}𝑑x={\int }_1^2{e}^{x^{-1}}{x}^{-2}𝑑x.$$

Let $u=x^{-1}$, the exponent on $e$. Then

	$du$	$=-x^{-2}dx$
	$-du$	$=x^{-2}dx.$

Bringing the negative sign outside the integral sign, the problem now reads

$$-\int e^u𝑑u.$$

Next, change the limits of integration:

	$u={(1)}^{-1}=1$
	$u={(2)}^{-1}={\displaystyle \frac{1}{2}}.$

Notice that now the limits begin with the larger number, meaning we must multiply by $-$1 and interchange the limits. Thus,

	$-{\displaystyle {\int }_1^{1/2}}{e}^u𝑑u$	$={\displaystyle {\int }_{1/2}^1}{e}^u𝑑u$
		$={e^u|}_{1/2}^1$
		$=e-e^{1/2}$
		$=e-\sqrt{e}.$

Solution: First factor the 3 outside the integral symbol. Then use the $u^{-1}$ rule. Thus,

	${\displaystyle \int \frac{3}{x-10}𝑑x}$	$=3{\displaystyle \int \frac{1}{x-10}𝑑x}$
		$=3{\displaystyle \int \frac{du}{u}}$
		$=3\ln |u|+C$
		$=3\ln |x-10|+C,x\ne 10.$

See Figure 5.42.

Solution: We need substitution to evaluate this problem. Let $u=1+\cos x,$, so $du=-\sin xdx$. Rewrite the integral in terms of $u$, changing the limits of integration as well. Thus,

	$u=1+\cos (0)=2$
	$u=1+\cos \left({\displaystyle \frac{\pi }{2}}\right)=1.$

Then

	${\displaystyle {\int }_0^{\pi /2}}{\displaystyle \frac{\sin x}{1+\cos x}}$	$=-{\displaystyle {\int }_2^1}{u}^{-1}𝑑u$
		$={\displaystyle {\int }_1^2}{u}^{-1}𝑑u$
		$={\ln |u||}_1^2$
		$=[\ln 2-\ln 1]$
		$=\ln 2.$

Solution: This can be rewritten as ${\displaystyle \int (2x^3+3x){(x^4+3x^2)}^{-1}𝑑x}$. Use substitution. Let $u=x^4+3x^2$, then $du=4x^3+6x$. Alter $du$ by factoring out the 2. Thus,

	$du$	$=(4x^3+6x)dx$
		$=2(2x^3+3x)dx$
	${\displaystyle \frac{1}{2}}du$	$=(2x^3+3x)dx.$

Rewrite the integrand in $u$:

$$\int (2x^3+3x){(x^4+3x^2)}^{-1}𝑑x=\frac{1}{2}\int u^{-1}𝑑u.$$

Then we have

	${\displaystyle \frac{1}{2}}{\displaystyle \int u^{-1}𝑑u}$	$={\displaystyle \frac{1}{2}}\ln |u|+C$
		$={\displaystyle \frac{1}{2}}\ln |x^4+3x^2|+C.$

Solution: Let us first use a trigonometric identity to rewrite the integral. The trig identity ${\cos }^2\theta ={\displaystyle \frac{1+\cos 2\theta }{2}}$ allows us to rewrite the integral as

$${\int }_0^{\pi /2}{\cos }^2\theta d\theta ={\int }_0^{\pi /2}\frac{1+\cos 2\theta }{2}𝑑\theta .$$

Then,

	${\displaystyle {\int }_0^{\pi /2}}\left({\displaystyle \frac{1+\cos 2\theta }{2}}\right)𝑑\theta$	$={\displaystyle {\int }_0^{\pi /2}}\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}\cos 2\theta \right)𝑑\theta$
		$={\displaystyle \frac{1}{2}}{\displaystyle {\int }_0^{\pi /2}}𝑑\theta +{\displaystyle \frac{1}{2}}{\displaystyle {\int }_0^{\pi /2}}\cos 2\theta d\theta .$

We can evaluate the first integral as it is, but we need to make a substitution to evaluate the second integral. Let $u=2\theta$. Then, $du=2d\theta$, or ${\displaystyle \frac{1}{2}}du=d\theta$. Also, when $\theta =0,u=0$, and when $\theta =\pi /2,u=\pi$. Expressing the second integral in terms of $u$, we have

	${\displaystyle \frac{1}{2}}{\displaystyle {\int }_0^{\pi /2}}𝑑\theta +{\displaystyle \frac{1}{2}}{\displaystyle {\int }_0^{\pi /2}}\cos 2\theta d\theta$	$={\displaystyle \frac{1}{2}}{\displaystyle {\int }_0^{\pi /2}}𝑑\theta +{\displaystyle \frac{1}{2}}\left({\displaystyle \frac{1}{2}}\right){\displaystyle {\int }_0^{\pi }}\cos udu$
		$={{\displaystyle \frac{\theta }{2}}|}_{\theta =0}^{\theta =\pi /2}+{{\displaystyle \frac{1}{4}}\sin u|}_{u=0}^{u=\theta }$
		$=\left({\displaystyle \frac{\pi }{4}}-0\right)+(0-0)={\displaystyle \frac{\pi }{4}}.$

Solution: We can go directly to the formula for the antiderivative in the rule on integration formulas resulting in inverse trigonometric functions, and then evaluate the definite integral. We have

	${\displaystyle {\int }_0^{\frac{1}{2}}}{\displaystyle \frac{dx}{\sqrt{1-x^2}}}$	$={{\sin }^{-1}x|}_0^{\frac{1}{2}}$
		$={\sin }^{-1}{\displaystyle \frac{1}{2}}-{\sin }^{-1}0$
		$={\displaystyle \frac{\pi }{6}}-0$
		$={\displaystyle \frac{\pi }{6}}.$

Solution: Substitute $u=3x$. Then $du=3dx$ and we have

$$\int \frac{dx}{\sqrt{4-9x^2}}=\frac{1}{3}\int \frac{du}{\sqrt{4-u^2}}.$$

Applying the formula with $a=2$, we obtain

	${\displaystyle \int \frac{dx}{\sqrt{4-9x^2}}}$	$={\displaystyle \frac{1}{3}}{\displaystyle \int \frac{du}{\sqrt{4-u^2}}}$
		$={\displaystyle \frac{1}{3}}{\sin }^{-1}\left({\displaystyle \frac{u}{2}}\right)+C$
		$={\displaystyle \frac{1}{3}}{\sin }^{-1}\left({\displaystyle \frac{3x}{2}}\right)+C.$

Solution: The format of the problem matches the inverse sine formula. Thus,

	${\displaystyle {\int }_0^{\sqrt{3}/2}}{\displaystyle \frac{du}{\sqrt{1-u^2}}}$	$={{\sin }^{-1}u|}_0^{\sqrt{3}/2}$
		$=\left[{\sin }^{-1}\left({\displaystyle \frac{\sqrt{3}}{2}}\right)\right]-\left[{\sin }^{-1}(0)\right]$
		$={\displaystyle \frac{\pi }{3}}.$

Solution: Comparing this problem with the formulas stated in the rule on integration formulas resulting in inverse trigonometric functions, the integrand looks similar to the formula for ${\tan }^{-1}u+C$. So we use substitution, letting $u=2x$, then $du=2dx$ and $1/2du=dx$. Then, we have

$$\frac{1}{2}\int \frac{1}{1+u^2}𝑑u=\frac{1}{2}{\tan }^{-1}u+C=\frac{1}{2}{\tan }^{-1}\left(2x\right)+C.$$

Solution: Apply the formula with $a=3$. Then,

$$\int \frac{dx}{9+x^2}=\frac{1}{3}{\tan }^{-1}\left(\frac{x}{3}\right)+C.$$