CE Board Exam Randomizer

Question 1

Trigonometric Integrals and Substitution

Two related but distinct tools live here. Trig integrals use identities to simplify powers of sin/cos before integrating: for odd powers, save one factor and use Pythagorean substitution ($\sin^2x=1-\cos^2x$); for even powers, use power-reduction ($\sin^2x=\tfrac{1-\cos2x}{2}$) to lower the degree. Trig substitution is a technique to eliminate square roots of the form $\sqrt{a^2\pm x^2}$ or $\sqrt{x^2-a^2}$ by swapping in a trig function whose Pythagorean identity collapses the radical. After integrating in $\theta$, always draw a right triangle from the original substitution to convert back to $x$. If the expression isn't in standard form yet, complete the square first.

Form under radical	Substitution	Simplification
$\sqrt{a^2-x^2}$	$x=a\sin\theta$	$\sqrt{a^2-x^2}=a\cos\theta$
$\sqrt{a^2+x^2}$	$x=a\tan\theta$	$\sqrt{a^2+x^2}=a\sec\theta$
$\sqrt{x^2-a^2}$	$x=a\sec\theta$	$\sqrt{x^2-a^2}=a\tan\theta$

Answer

Using the beta/Wallis result,

$$\int_0^{\pi/2}\sin^6x\cos^4x\,dx=\frac{1}{2}B\left(\frac{7}{2},\frac{5}{2}\right)=\frac{3\pi}{512}$$

Therefore,

$$\int_0^{\pi/2}2\sin^6x\cos^4x\,dx=\frac{3\pi}{256}=\boxed{0.0368}$$

Answer

For an even power,

$$\int_0^{\pi/2}\cos^8x\,dx=\frac{7\cdot5\cdot3\cdot1}{8\cdot6\cdot4\cdot2}\cdot\frac{\pi}{2}$$

$$=\frac{35\pi}{256}=\boxed{0.4295}$$

Answer

On $[0,\pi/4]$, $\cos x$ is above $\sin x$.

$$A=\int_0^{\pi/4}(\cos x-\sin x)\,dx=\left[\sin x+\cos x\right]_0^{\pi/4}$$

$$A=\sqrt{2}-1=\boxed{0.414}$$

Answer

The curves meet at $x=\pi/4$. The region equals the area under $y=\cos x$ from $0$ to $\pi/2$ minus the area between $\cos x$ and $\sin x$ from $0$ to $\pi/4$.

$$A=\int_0^{\pi/2}\cos x\,dx-(\sqrt{2}-1)=1-\sqrt{2}+1$$

$$A=2-\sqrt{2}=\boxed{0.586\text{ sq. units}}$$

Answer

Use $x=5\sin\theta$, so $dx=5\cos\theta\,d\theta$ and $\sqrt{25-x^2}=5\cos\theta$.

$$\int\frac{dx}{\sqrt{25-x^2}}=\int d\theta=\theta+C$$

Since $\theta=\sin^{-1}(x/5)$,

$$\boxed{\sin^{-1}\left(\frac{x}{5}\right)+C}$$

Answer

Let $x=2\sin\theta$, $dx=2\cos\theta\,d\theta$, $\sqrt{4-x^2}=2\cos\theta$.

$$\int x^2\sqrt{4-x^2}\,dx=\int(2\sin\theta)^2(2\cos\theta)(2\cos\theta\,d\theta)=16\int\sin^2\theta\cos^2\theta\,d\theta$$

Use $\sin^2\theta\cos^2\theta=\frac{1}{4}\sin^2(2\theta)=\frac{1-\cos4\theta}{8}$:

$$=16\int\frac{1-\cos4\theta}{8}\,d\theta=2\left(\theta-\frac{\sin4\theta}{4}\right)+C$$

Back-substitute using the triangle from $x=2\sin\theta$: $\sin\theta=x/2$, $\cos\theta=\sqrt{4-x^2}/2$, so $\sin2\theta=2\sin\theta\cos\theta=\frac{x\sqrt{4-x^2}}{2}$ and $\sin4\theta=2\sin2\theta\cos2\theta$. After simplification:

$$\boxed{\frac{x(2x^2-4)\sqrt{4-x^2}}{8}+\frac{1}{2}\sin^{-1}\!\left(\frac{x}{2}\right)+C}$$

Answer

Let $x=3\tan\theta$, $dx=3\sec^2\theta\,d\theta$, $(x^2+9)^{3/2}=(9\sec^2\theta)^{3/2}=27\sec^3\theta$.

$$\int\frac{dx}{(x^2+9)^{3/2}}=\int\frac{3\sec^2\theta\,d\theta}{27\sec^3\theta}=\frac{1}{9}\int\cos\theta\,d\theta=\frac{\sin\theta}{9}+C$$

Back-substitute: from $x=3\tan\theta$, draw a right triangle with opposite $=x$, adjacent $=3$, hypotenuse $=\sqrt{x^2+9}$. So $\sin\theta=\frac{x}{\sqrt{x^2+9}}$.

$$\boxed{\frac{x}{9\sqrt{x^2+9}}+C}$$

Answer

Let $x=2\sec\theta$, $dx=2\sec\theta\tan\theta\,d\theta$, $\sqrt{x^2-4}=2\tan\theta$.

$$\int\frac{2\tan\theta}{2\sec\theta}\cdot2\sec\theta\tan\theta\,d\theta=2\int\tan^2\theta\,d\theta=2\int(\sec^2\theta-1)\,d\theta$$

$$=2(\tan\theta-\theta)+C$$

Back-substitute: $\sec\theta=x/2$, so $\tan\theta=\frac{\sqrt{x^2-4}}{2}$ and $\theta=\sec^{-1}(x/2)$.

$$\boxed{\sqrt{x^2-4}-2\sec^{-1}\!\left(\frac{x}{2}\right)+C}$$

Answer

Step 1 — Complete the square in the radicand: $-x^2+4x=-(x^2-4x)=-(x^2-4x+4-4)=4-(x-2)^2$.

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

Step 2: Let $u=x-2$, so $du=dx$:

$$\int\frac{du}{\sqrt{4-u^2}}=\sin^{-1}\!\left(\frac{u}{2}\right)+C$$

Back-substitute $u=x-2$:

$$\boxed{\sin^{-1}\!\left(\frac{x-2}{2}\right)+C}$$

Answer

Let $x=2\sin\theta$, $dx=2\cos\theta\,d\theta$, $\sqrt{4-x^2}=2\cos\theta$. Change limits: $x=0\Rightarrow\theta=0$; $x=2\Rightarrow\theta=\pi/2$.

$$\int_0^{\pi/2}\frac{8\sin^3\theta}{2\cos\theta}\cdot2\cos\theta\,d\theta=8\int_0^{\pi/2}\sin^3\theta\,d\theta$$

Split $\sin^3\theta=\sin\theta(1-\cos^2\theta)$ and let $u=\cos\theta$:

$$=8\int_1^0-(1-u^2)\,du=8\int_0^1(1-u^2)\,du=8\left[u-\frac{u^3}{3}\right]_0^1=8\cdot\frac{2}{3}$$

$$\boxed{\frac{16}{3}}$$

Answer

Arc length formula: $L=\int_0^1\sqrt{1+(y')^2}\,dx$. Since $y'=x$:

$$L=\int_0^1\sqrt{1+x^2}\,dx$$

Let $x=\tan\theta$, $dx=\sec^2\theta\,d\theta$, $\sqrt{1+x^2}=\sec\theta$. Limits: $\theta=0$ to $\theta=\pi/4$.

$$L=\int_0^{\pi/4}\sec^3\theta\,d\theta=\left[\frac{\sec\theta\tan\theta}{2}+\frac{\ln|\sec\theta+\tan\theta|}{2}\right]_0^{\pi/4}$$

$$=\frac{\sqrt{2}\cdot1}{2}+\frac{\ln(\sqrt{2}+1)}{2}=\frac{\sqrt{2}}{2}+\frac{\ln(\sqrt{2}+1)}{2}$$

$$\boxed{L\approx1.1478}$$

Answer

For $y=x^{2/3}$:
$y'=\frac{2}{3}x^{-1/3}$
Let $u=x^{1/3}$, so $x=u^3$ and $dx=3u^2du$. Arc length:
$L=\int_0^8\sqrt{1+(y')^2}dx=\int_0^2 u\sqrt{9u^2+4}\,du$
$L=\frac{(9u^2+4)^{3/2}}{27}\Big|_0^2$
$\boxed{9.073}$

Answer

The second plan is more profitable while
$$R_2(x)-R_1(x)>0$$
$$200+5x-(50+x^2)=150+5x-x^2>0$$
Set equal to zero:
$$x^2-5x-150=0$$
The positive root is $x=15$, so plan 2 is better for 15 years. Net excess profit:
$$\int_0^{15}(150+5x-x^2)dx$$
$$=\left[150x+\frac{5x^2}{2}-\frac{x^3}{3}\right]_0^{15}=1687.50$$
Therefore, the answer is $\boxed{15\text{ yrs},\ P1687.50}$.

Answer

Net savings rate is
$$S(x)-C(x)=150-x^2-\left(x^2+\frac{11x}{4}\right)=150-2x^2-\frac{11x}{4}$$
During the first year, integrate from 0 to 1:
$$\int_0^1\left(150-2x^2-\frac{11x}{4}\right)dx$$
$$=\left[150x-\frac{2x^3}{3}-\frac{11x^2}{8}\right]_0^1$$
$$=150-0.667-1.375=147.96$$
Therefore, the first-year net savings are approximately $\boxed{148}$.

Answer

Use the Beta integral form:
$$\int_0^{\pi/2}\sin^6\phi\cos^4\phi\,d\phi=\frac{1}{2}B\left(\frac{7}{2},\frac{5}{2}\right)$$
This evaluates to
$$\frac{45\pi}{7680}=0.0184$$
Therefore, the value is $\boxed{0.0184}$.

Answer

Interpreting the intended limits as $0$ to $\pi/2$, use the Beta integral form:
$$\int_0^{\pi/2}\sin^5\phi\cos^7\phi\,d\phi=\frac{1}{2}B\left(3,4\right)$$
$$=\frac{1}{2}\left(\frac{2!3!}{6!}\right)=\frac{1}{120}$$
$$\frac{1}{120}=0.0083$$
Therefore, the value is $\boxed{0.0083}$.

Answer

Using Wallis' formula for even powers,
$$\int_0^{\pi/2}\sin^4x\,dx=\frac{3\cdot1}{4\cdot2}\cdot\frac{\pi}{2}=\frac{3\pi}{16}$$
$$\frac{3\pi}{16}=0.589$$
Therefore, the value is $\boxed{0.589}$.

Answer

Using Wallis' formula for odd powers,
$$\int_0^{\pi/2}\cos^5\phi\,d\phi=\frac{4\cdot2}{5\cdot3\cdot1}=\frac{8}{15}$$
$$\frac{8}{15}=0.533$$
Therefore, the value is $\boxed{0.533}$.

Answer

Let $u=3A$, so $dA=du/3$. The limits change from $A=0$ to $u=0$, and from $A=\pi/6$ to $u=\pi/2$. Thus,
$$\int_0^{\pi/6}\cos^8(3A)\,dA=\frac{1}{3}\int_0^{\pi/2}\cos^8u\,du$$
For even powers,
$$\int_0^{\pi/2}\cos^8u\,du=\frac{7\cdot5\cdot3\cdot1}{8\cdot6\cdot4\cdot2}\cdot\frac{\pi}{2}=\frac{105\pi}{768}$$
Therefore,
$$\frac{1}{3}\left(\frac{105\pi}{768}\right)=\frac{35\pi}{768}$$
So the value is $\boxed{35\pi/768}$.

Answer

Use the Beta integral form:
$$\int_0^{\pi/2}\sin^5x\cos^3x\,dx=\frac{1}{2}B\left(\frac{6}{2},\frac{4}{2}\right)$$
$$=\frac{1}{2}B(3,2)=\frac{1}{2}\left(\frac{2!1!}{4!}\right)=\frac{1}{24}$$
$$\frac{1}{24}=0.0417$$
Therefore, the value is $\boxed{0.0417}$.

Answer

Using Wallis' formula for odd powers,
$$\int_0^{\pi/2}\sin^7x\,dx=\frac{6\cdot4\cdot2}{7\cdot5\cdot3\cdot1}=\frac{16}{35}$$
Thus,
$$15\int_0^{\pi/2}\sin^7x\,dx=15\left(\frac{16}{35}\right)=6.857$$
Therefore, the value is $\boxed{6.857}$.

Answer

Use the Wallis/Beta integral form:
$$\int_0^{\pi/2}\sin^m x\cos^n x\,dx=\frac{1}{2}B\left(\frac{m+1}{2},\frac{n+1}{2}\right)$$
Here $m=2$ and $n=6$, so
$$5\int_0^{\pi/2}\sin^2x\cos^6x\,dx=5\left[\frac{1}{2}B\left(\frac{3}{2},\frac{7}{2}\right)\right]$$
$$=0.3068$$
Therefore, the value is $\boxed{0.307}$.

Answer

Evaluate
$$3\int_0^{\pi/2}\sin^3x\,dx$$
Using Wallis' result,
$$\int_0^{\pi/2}\sin^3x\,dx=\frac{2}{3}$$
Therefore,
$$3\left(\frac{2}{3}\right)=2$$
Thus, the value of the integral is $\boxed{2}$.

Question 2

Product-to-Sum Identities

$$2\sin A\sin B=\cos(A-B)-\cos(A+B)$$

$$2\cos A\cos B=\cos(A-B)+\cos(A+B)$$

$$2\sin A\cos B=\sin(A-B)+\sin(A+B)$$

Power-Reduction and Pythagorean Identities

$$\sin^2 x=\frac{1-\cos 2x}{2}$$

$$\cos^2 x=\frac{1+\cos 2x}{2}$$

$$1+\tan^2 x=\sec^2 x$$

$$1+\cot^2 x=\csc^2 x$$

Basic Trigonometric Integrals

$$\int \sin u\,du = -\cos u + C$$

$$\int \cos u\,du = \sin u + C$$

$$\int \sec^2 u\,du = \tan u + C$$

$$\int \csc^2 u\,du = -\cot u + C$$

$$\int \sec u\,\tan u\,du = \sec u + C$$

$$\int \csc u\,\cot u\,du = -\csc u + C$$

$$\int \tan u\,du = \ln|\sec u| + C$$

$$\int \cot u\,du = \ln|\sin u| + C$$

$$\int \sec u\,du = \ln|\sec u+\tan u| + C$$

$$\int \csc u\,du = \ln|\csc u-\cot u| + C$$

Inverse Trigonometric Forms

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

$$\int \frac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\!\left(\frac{u}{a}\right)+C$$

$$\int \frac{du}{a^2-u^2}=\frac{1}{2a}\ln\left|\frac{a+u}{a-u}\right|+C$$

$$\int \frac{du}{\sqrt{u^2-a^2}}=\ln\left|u+\sqrt{u^2-a^2}\right|+C$$

Trigonometric Substitution

Case 1: Expressions of the Form $\sqrt{a^2 - x^2}$

$$x = a\sin\theta$$

$$dx = a\cos\theta\,d\theta$$

$$\sqrt{a^2 - x^2} = a\cos\theta$$

Case 2: Expressions of the Form $\sqrt{a^2 + x^2}$

$$x = a\tan\theta$$

$$dx = a\sec^2\theta\,d\theta$$

$$\sqrt{a^2 + x^2} = a\sec\theta$$

Case 3: Expressions of the Form $\sqrt{x^2 - a^2}$

$$x = a\sec\theta$$

$$dx = a\sec\theta\tan\theta\,d\theta$$

$$\sqrt{x^2 - a^2} = a\tan\theta$$

Back-Substitution Using Right Triangles

$$\sin\theta=\frac{x}{a}\qquad \cos\theta=\frac{\sqrt{a^2-x^2}}{a}$$

$$\tan\theta=\frac{x}{a}\qquad \sec\theta=\frac{\sqrt{a^2+x^2}}{a}$$

$$\sec\theta=\frac{x}{a}\qquad \tan\theta=\frac{\sqrt{x^2-a^2}}{a}$$

Question 3

Wallis Formula for Sine and Cosine Powers

Evaluate $$\int_0^{\pi/2}2\sin^6x\cos^4x\,dx.$$

Question 4

Even-Power Cosine Integral

Evaluate $$\int_0^{\pi/2}\cos^8x\,dx.$$

Question 5

Area Between Sine and Cosine on a Short Interval

Find the area between $y=\sin x$ and $y=\cos x$ from $x=0$ to $x=\pi/4$.

Question 6

First-Quadrant Area With Sine, Cosine, and Axis

Find the area bounded by $y=\sin x$, $y=\cos x$, and the $x$-axis in the first quadrant.

Question 7

Trigonometric Substitution for Circle Form

Evaluate $$\int\frac{dx}{\sqrt{25-x^2}}.$$

Question 8

Case 1 Full Back-Substitution: $\sqrt{a^2-x^2}$

Evaluate $$\int x^2\sqrt{4-x^2}\,dx.$$

Question 9

Case 2 Full Example: $\sqrt{a^2+x^2}$

Evaluate $$\int\frac{dx}{(x^2+9)^{3/2}}.$$

Question 10

Case 3 Full Example: $\sqrt{x^2-a^2}$

Evaluate $$\int\frac{\sqrt{x^2-4}}{x}\,dx.$$

Question 11

Completing the Square Before Trig Sub

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

Question 12

Definite Integral with Trig Substitution

Evaluate $$\int_0^{2}\frac{x^3}{\sqrt{4-x^2}}\,dx.$$

Question 13

Arc Length Using Trig Sub

Find the arc length of the parabola $y=\tfrac{x^2}{2}$ from $x=0$ to $x=1$.

Question 14

Question Bank: q468

MSTE - Integral Calculus / Arc Length / Engr. Janclyde Espinosa (Clidez)

The length of arc of the function f(x)=x^2/3 from x=0 to x=8 is:

Answer:

9.073
8.765
10.231
12.988

Question 15

Question Bank: t1668

MSTE - Integral Calculus / Applications of Integration / BEMz

In x years from now, one investment plan will be generating profit at the rate of $R1(x)=50+x^{2}$ pesos per yr, while a second plan will be generating profit at the rate $R2(x)=200+5x$ pesos per yr. For how many yrs will the second plan be more profitable one? Compute also the net excess profit if the second plan would be used instead of the first.

15yrs, P1687.50
12yrs, P1450.25
14yrs, P15640.25
10yrs, P1360.25

Question 16

Question Bank: t1679

MSTE - Integral Calculus / Applications of Integration / BEMz

Suppose a company wants to introduce a new machine that will produce a rate of annual savings $S(x)=150-x^{2}$ where x is the number of yrs of operation of the machine, while producing a rate of annual costs of $C(x)=(x^{2})+(11x/4)$. What are the net total savings during the first year of use of the machine?

122
148
257
183

Question 17