Calculus 2: Integration

/26

Test your understanding of integration techniques in Calculus 2, including substitution, integration by parts, partial fractions, and applications of definite integrals.

1 / 26

The derivative of

$f(x) = sin^{-1}(3x)$

is:

frac{1}{sqrt{1 – 9x^2}}

frac{3}{sqrt{1 – x^2}}

frac{3}{sqrt{1 – 9x^2}}

frac{1}{sqrt{1 – 3x^2}}

Use the chain rule:

$frac{d}{dx} sin^{-1}(u) = frac{u’}{sqrt{1 – u^2}}$

, where

$u = 3x$

2 / 26

$y = ln(sec x + tan x)$

, then

$frac{dy}{dx}$

is:

sec x

tan x

sec^2 x

csc x

3 / 26

The derivative of

$f(x) = x^x$

is:

x^x ln x

x^{x-1}

e^{x ln x}

x^x (1 + ln x)

Rewrite

$x^x = e^{x ln x}$

, then differentiate using the chain rule.

4 / 26

The derivative of

$f(x) = tanh^{-1}(x)$

is:

frac{1}{1 + x^2}

frac{1}{1 – x^2}

frac{1}{sqrt{1 – x^2}}

frac{1}{sqrt{1 + x^2}}

5 / 26

$y = e^{x sin x}$

, then

$frac{dy}{dx}$

is:

e^{x sin x} (sin x + x cos x)

e^{x sin x} cos x

e^{x sin x} sin x

e^{x sin x} (x sin x)

6 / 26

The derivative of

$f(x) = frac{ln x}{x^2}$

is:

frac{1 + 2 ln x}{x^3}

frac{2 ln x – 1}{x^3}

frac{ln x – 1}{x^3}

frac{1 – 2 ln x}{x^3}

Use the quotient rule:

$frac{d}{dx} left( frac{u}{v} right) = frac{u’v – uv’}{v^2}$

7 / 26

$f(x) = sqrt{x + sqrt{x}}$

, then

$f'(x)$

is:

frac{1}{2 sqrt{x + sqrt{x}}}

frac{1 + frac{1}{2sqrt{x}}}{2 sqrt{x + sqrt{x}}}

frac{1}{4 sqrt{x} sqrt{x + sqrt{x}}}

frac{1}{2 sqrt{x}}

8 / 26

The derivative of

$f(x) = sec^{-1}(x^2)$

is:

frac{2x}{

sqrt{x^4 – 1}}

frac{1}{

Use the chain rule and derivative of

sec^{-1}(u)

9 / 26

$y = cosh(2x)$

, then

$frac{d^2y}{dx^2}$

is:

4 cosh(2x)

2 sinh(2x)

4 sinh(2x)

cosh(2x)

First derivative is

$2 sinh(2x)$

, second derivative is

$4 cosh(2x)$

10 / 26

$x = t^2$

and

$y = t^3$

, then

$frac{dy}{dx}$

is:

frac{3}{2t}

frac{3t}{2}

frac{2}{3t}

frac{2t}{3}

$frac{dy}{dx} = frac{dy/dt}{dx/dt} = frac{3t^2}{2t} = frac{3t}{2}$

11 / 26

For the curve

$x^2 + y^2 = 25$

$frac{dy}{dx}$

is:

frac{x}{y}

-frac{y}{x}

frac{y}{x}

-frac{x}{y}

Differentiate implicitly:

$2x + 2y frac{dy}{dx} = 0$

12 / 26

$y = sin(xy)$

, then

$frac{dy}{dx}$

is:

frac{cos(xy)}{1 – x cos(xy)}

frac{y cos(xy)}{1 – x cos(xy)}

frac{y cos(xy)}{x cos(xy)}

frac{cos(xy)}{x cos(xy)}

Differentiate implicitly and solve for

$frac{dy}{dx}$

13 / 26

For parametric equations

$x = e^t$

$y = t e^t$

$frac{d^2y}{dx^2}$

is:

frac{1}{e^t}

frac{t}{e^t}

frac{1 + t}{e^t}

frac{1 – t}{e^t}

First find

$frac{dy}{dx} = frac{1 + t}{e^t}$

, then differentiate again.

14 / 26

The slope of the tangent to the curve

$x^3 + y^3 = 9$

$(2, 1)$

is:

2

-1

-2

1

Differentiate implicitly and substitute

$(2, 1)$

15 / 26

$y$

is defined implicitly by

$x^2 y + y^2 x = 6$

, then

$frac{dy}{dx}$

$(1, 2)$

is:

frac{6}{5}

-frac{5}{6}

frac{5}{6}

-frac{6}{5}

Differentiate implicitly and solve for

$frac{dy}{dx}$

16 / 26

For

$x = cos theta$

$y = sin theta$

$frac{d^2y}{dx^2}$

is:

-sec^3 theta

sec^3 theta

-csc^3 theta

csc^3 theta

Compute

$frac{dy}{dx} = -cot theta$

, then differentiate again.

17 / 26

The derivative

$frac{dy}{dx}$

for

$ln(xy) = x + y$

is:

frac{x(1 – y)}{y(x – 1)}

frac{y(1 – x)}{x(y – 1)}

frac{xy – 1}{xy + 1}

frac{xy + 1}{xy – 1}

Differentiate implicitly and solve for

$frac{dy}{dx}$

18 / 26

$x = t – sin t$

$y = 1 – cos t$

, then

$frac{dy}{dx}$

is:

frac{sin t}{1 – cos t}

frac{1 – cos t}{sin t}

frac{cos t}{sin t}

frac{sin t}{cos t}

$frac{dy}{dx} = frac{dy/dt}{dx/dt} = frac{sin t}{1 – cos t}$

19 / 26

The second derivative

$frac{d^2y}{dx^2}$

for

$y^2 = 4x$

is:

-frac{1}{2y^3}

frac{1}{2y^3}

-frac{1}{y^3}

frac{1}{y^3}

20 / 26

The derivative $frac{dy}{dx}$ for $y = x^{x^x}$ is best found using:

Logarithmic differentiation

Implicit differentiation

Chain rule

Quotient rule

21 / 26

The family of curves orthogonal to

$y = Cx^2$

is:

y^2 + 2x^2 = K

x^2 + 2y^2 = K

x^2 – y^2 = K

xy = K

Orthogonal trajectories satisfy

$frac{dy}{dx} = -frac{1}{f'(x)}$

22 / 26

The derivative of

$y = (sin x)^{ln x}$

is:

(sin x)^{ln x} left( frac{ln x}{x} + cot x right)

(sin x)^{ln x} ln(sin x)

(sin x)^{ln x} left( frac{ln(sin x)}{x} + ln x cot x right)

(sin x)^{ln x} ln x cot x

23 / 26

The derivative of

$f(x) = int_0^x e^{t^2} dt$

is:

e^{x^2}

e^{x^2} – 1

2x e^{x^2}

frac{e^{x^2}}{2x}

By the Fundamental Theorem of Calculus,

$frac{d}{dx} int_a^x f(t) dt = f(x)$

24 / 26

The derivative of

$y = tan^{-1}(sinh x)$

is:

text{sech} x

text{csch} x

tanh x

cosh x

Use the chain rule and derivative of

$tan^{-1}(u)$

25 / 26

The derivative of

$f(x) = frac{x}{sqrt{a^2 – x^2}}$

is:

frac{a^2}{(a^2 – x^2)^{3/2}}

frac{x^2}{(a^2 – x^2)^{3/2}}

frac{1}{sqrt{a^2 – x^2}}

frac{a^2 – 2x^2}{(a^2 – x^2)^{3/2}}

26 / 26

The derivative of

$f(x) = frac{e^x – e^{-x}}{e^x + e^{-x}}$

is:

1 – f(x)^2

1 + f(x)^2

f(x)^2 – 1

-f(x)^2

Recognize

$f(x) = tanh x$

, whose derivative is

$text{sech}^2 x = 1 – tanh^2 x$

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