Calculus Study Summary

This summary covers various topics in differential and integral calculus, as presented in the provided multiple-choice questions.

I. Integration Techniques

1. Integration by Substitution

• Problem: Evaluate $\int \frac{\cos(\ln x)}{x} dx$
  Solution: Let $u = \ln x$. Then $du = \frac{1}{x} dx$. The integral becomes $\int \cos(u) du = \sin(u) + C$. Substituting back, we get $\sin(\ln x) + C$.
  Answer: $\sin(\ln x) + C$
• Problem: Simplify $\int 3x^2 e^{x^3} dx$
  Solution: Let $u = x^3$. Then $du = 3x^2 dx$. The integral becomes $\int e^u du = e^u + C$. Substituting back, we get $e^{x^3} + C$.
  Answer: $e^{x^3} + C$
• Problem: Evaluate $\int \frac{e^{2x} + e^{-2x}}{e^{2x} - e^{-2x}} dx$
  Solution: Let $u = e^{2x} - e^{-2x}$. Then $du = (2e^{2x} - (-2e^{-2x})) dx = 2(e^{2x} + e^{-2x}) dx$. So, $\frac{1}{2} du = (e^{2x} + e^{-2x}) dx$. The integral becomes $\int \frac{1}{u} \cdot \frac{1}{2} du = \frac{1}{2} \ln|u| + C$. Substituting back, we get $\frac{1}{2} \ln|e^{2x} - e^{-2x}| + C$.
  Answer: $\frac{1}{2} \ln|e^{2x} - e^{-2x}| + C$
• Problem: Evaluate $\int \cos^3 x dx$
  Solution: Rewrite $\cos^3 x$ as $\cos^2 x \cdot \cos x = (1 - \sin^2 x) \cos x$. The integral becomes $\int (1 - \sin^2 x) \cos x dx$. Let $u = \sin x$. Then $du = \cos x dx$. The integral becomes $\int (1 - u^2) du = u - \frac{u^3}{3} + C$. Substituting back, we get $\sin x - \frac{\sin^3 x}{3} + C$.
  Answer: $\sin x - \frac{1}{3} \sin^3 x + C$

2. Integration by Partial Fractions

• Problem: Simplify $\int \frac{4x-5}{x^2-x-2} dx$
  Solution: Factor the denominator: $x^2-x-2 = (x-2)(x+1)$. Perform partial fraction decomposition: $\frac{4x-5}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1}$. Solving for A and B: $4x-5 = A(x+1) + B(x-2)$. Setting $x=2 \implies 3 = 3A \implies A=1$. Setting $x=-1 \implies -9 = -3B \implies B=3$. The integral becomes $\int \left( \frac{1}{x-2} + \frac{3}{x+1} \right) dx = \ln|x-2| + 3\ln|x+1| + C$.
  Answer: $3\ln|x+1| + \ln|x-2| + C$

3. Integration by Parts

• Problem: Simplify $\int e^x \sin x dx$
  Solution: Use integration by parts twice: $\int u dv = uv - \int v du$. Let $I = \int e^x \sin x dx$. 1. Let $u = \sin x$, $dv = e^x dx$. Then $du = \cos x dx$, $v = e^x$. $I = e^x \sin x - \int e^x \cos x dx$. 2. For $\int e^x \cos x dx$: Let $u = \cos x$, $dv = e^x dx$. Then $du = -\sin x dx$, $v = e^x$. $\int e^x \cos x dx = e^x \cos x - \int e^x (-\sin x) dx = e^x \cos x + \int e^x \sin x dx$. Substitute back into $I$: $I = e^x \sin x - (e^x \cos x + I)$. $I = e^x \sin x - e^x \cos x - I$. $2I = e^x (\sin x - \cos x)$. $I = \frac{e^x}{2} (\sin x - \cos x) + C$.
  Answer: $\frac{e^x}{2} (\sin x - \cos x) + C$
• Problem: Simplify $\int \ln x dx$
  Solution: Use integration by parts: $\int u dv = uv - \int v du$. Let $u = \ln x$, $dv = dx$. Then $du = \frac{1}{x} dx$, $v = x$. Integral: $x \ln x - \int x \cdot \frac{1}{x} dx = x \ln x - \int 1 dx = x \ln x - x + C$.
  Answer: $x \ln x - x + C$

II. Differentiation Techniques

1. Differentiation of Logarithmic Functions

• Problem: Find $\frac{dy}{dx}$ if $y = \log_e (\cos x)$
  Solution: Using the chain rule, $\frac{d}{dx} (\ln f(x)) = \frac{f'(x)}{f(x)}$. Here $f(x) = \cos x$, so $f'(x) = -\sin x$. $\frac{dy}{dx} = \frac{-\sin x}{\cos x} = -\tan x$.
  Answer: $-\tan x$
• Problem: Obtain $\frac{d}{dx} \left\{ \ln \left( \frac{x^2}{x^2+1} \right) \right\}$
  Solution: First, use logarithm properties: $\ln \left( \frac{x^2}{x^2+1} \right) = \ln(x^2) - \ln(x^2+1) = 2\ln x - \ln(x^2+1)$. Now differentiate: $\frac{d}{dx} (2\ln x - \ln(x^2+1)) = 2 \cdot \frac{1}{x} - \frac{1}{x^2+1} \cdot (2x)$ $= \frac{2}{x} - \frac{2x}{x^2+1} = \frac{2(x^2+1) - 2x(x)}{x(x^2+1)} = \frac{2x^2+2 - 2x^2}{x(x^2+1)} = \frac{2}{x(x^2+1)}$.
  Answer: $\frac{2}{x(x^2+1)}$
• Problem: Differentiate $y = \log(x^3 - x) + 7^{\frac{2}{3}x}$
  Solution: Assuming $\log$ refers to the natural logarithm $\ln$: 1. For $\ln(x^3 - x)$: $\frac{d}{dx} (\ln(x^3 - x)) = \frac{1}{x^3 - x} \cdot (3x^2 - 1) = \frac{3x^2 - 1}{x(x^2 - 1)}$. 2. For $7^{\frac{2}{3}x}$: Use $\frac{d}{dx} (a^{f(x)}) = a^{f(x)} \ln a \cdot f'(x)$. Here $a=7$, $f(x) = \frac{2}{3}x$, so $f'(x) = \frac{2}{3}$. $\frac{d}{dx} (7^{\frac{2}{3}x}) = 7^{\frac{2}{3}x} \ln 7 \cdot \frac{2}{3}$. Combining the results: $\frac{dy}{dx} = \frac{3x^2 - 1}{x(x^2 - 1)} + 7^{\frac{2}{3}x} \left(\frac{2}{3} \ln 7\right)$.
  Answer: $\frac{3x^2 - 1}{x(x^2 - 1)} + 7^{\frac{2}{3}x} \left(\frac{2}{3} \log 7\right)$

2. Product Rule

• Problem: If $y = e^{2x} \log(x+6)$, find $\frac{dy}{dx}$.
  Solution: Use the product rule: $\frac{d}{dx} (uv) = u'v + uv'$. Let $u = e^{2x}$, $u' = 2e^{2x}$. Let $v = \log(x+6)$, $v' = \frac{1}{x+6}$ (assuming $\log$ is $\ln$). $\frac{dy}{dx} = (2e^{2x}) \log(x+6) + e^{2x} \left(\frac{1}{x+6}\right)$. Factoring out $e^{2x}$: $\frac{dy}{dx} = e^{2x} \left( \frac{1}{x+6} + 2\log(x+6) \right)$.
  Answer: $e^{2x} \left[ \frac{1}{x+6} + 2\log(x+6) \right]$

3. Quotient Rule

• Problem: If $y = \frac{e^{\sin 2x}}{\cos 2x}$, find $\frac{dy}{dx}$.
  Solution: Use the quotient rule: $\frac{d}{dx} \left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$. Let $u = e^{\sin 2x}$, $u' = e^{\sin 2x} \cdot (\cos 2x) \cdot 2 = 2e^{\sin 2x} \cos 2x$. Let $v = \cos 2x$, $v' = -\sin 2x \cdot 2 = -2\sin 2x$. $\frac{dy}{dx} = \frac{(2e^{\sin 2x} \cos 2x)(\cos 2x) - (e^{\sin 2x})(-2\sin 2x)}{(\cos 2x)^2}$ $= \frac{2e^{\sin 2x} \cos^2 2x + 2e^{\sin 2x} \sin 2x}{\cos^2 2x}$ Factor out $2e^{\sin 2x}$: $2e^{\sin 2x} \left( \frac{\cos^2 2x + \sin 2x}{\cos^2 2x} \right)$ $= 2e^{\sin 2x} \left( 1 + \frac{\sin 2x}{\cos^2 2x} \right) = 2e^{\sin 2x} (1 + \sec 2x \tan 2x)$.
  Answer: $2e^{\sin 2x} [1 + \sec 2x \tan 2x]$

4. Implicit Differentiation

• Problem: Differentiate $x^2 \sin y - e^{2y} \cos x = 3x^2 + 2x - 5$ with respect to $x$.
  Solution: Apply $\frac{d}{dx}$ to each term: $\frac{d}{dx}(x^2 \sin y) - \frac{d}{dx}(e^{2y} \cos x) = \frac{d}{dx}(3x^2 + 2x - 5)$ Product Rule for $x^2 \sin y$: $2x \sin y + x^2 (\cos y) \frac{dy}{dx}$. Product Rule for $e^{2y} \cos x$: $(2e^{2y} \frac{dy}{dx}) \cos x + e^{2y} (-\sin x)$. Differentiating right side: $6x + 2$. Combining: $2x \sin y + x^2 \cos y \frac{dy}{dx} - (2e^{2y} \cos x \frac{dy}{dx} - e^{2y} \sin x) = 6x + 2$. $2x \sin y + x^2 \cos y \frac{dy}{dx} - 2e^{2y} \cos x \frac{dy}{dx} + e^{2y} \sin x = 6x + 2$. Group $\frac{dy}{dx}$ terms: $\frac{dy}{dx} (x^2 \cos y - 2e^{2y} \cos x) = 6x + 2 - 2x \sin y - e^{2y} \sin x$. $\frac{dy}{dx} = \frac{6x - 2x \sin y - e^{2y} \sin x + 2}{x^2 \cos y - 2e^{2y} \cos x}$.
  Answer: $\frac{6x - 2x \sin y - \sin x e^{2y} + 2}{x^2 \cos y - 2\cos x e^{2y}}$
• Problem: Find $\frac{dy}{dx}$ if $3x^2y + xy^2 = 5$.
  Solution: Differentiate each term with respect to $x$: $\frac{d}{dx} (3x^2y) + \frac{d}{dx} (xy^2) = \frac{d}{dx} (5)$ Product Rule for $3x^2y$: $(6x)y + 3x^2 \frac{dy}{dx}$. Product Rule for $xy^2$: $(1)y^2 + x (2y \frac{dy}{dx})$. Differentiating right side: $0$. Combining: $6xy + 3x^2 \frac{dy}{dx} + y^2 + 2xy \frac{dy}{dx} = 0$. Group $\frac{dy}{dx}$ terms: $(3x^2 + 2xy) \frac{dy}{dx} = -6xy - y^2$. $\frac{dy}{dx} = \frac{-6xy - y^2}{3x^2 + 2xy} = \frac{-(y^2 + 6xy)}{3x^2 + 2xy}$.
  Answer: $\frac{-(y^2+6xy)}{3x^2+2xy}$

5. Logarithmic Differentiation (for $f(x)^{g(x)}$ form)

• Problem: Find $\frac{dy}{dx}$ if $y = x^{\tan x}$.
  Solution: Take the natural logarithm of both sides: $\ln y = \ln (x^{\tan x}) = \tan x \cdot \ln x$. Differentiate implicitly with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (\tan x \cdot \ln x)$. Apply the product rule on the right side: $(\sec^2 x) \ln x + \tan x \cdot \frac{1}{x}$. So, $\frac{1}{y} \frac{dy}{dx} = \sec^2 x \ln x + \frac{\tan x}{x}$. Multiply by $y$: $\frac{dy}{dx} = y \left( \frac{\tan x}{x} + \sec^2 x \ln x \right)$. Substitute $y = x^{\tan x}$ back: $\frac{dy}{dx} = x^{\tan x} \left( \frac{\tan x}{x} + \sec^2 x \ln x \right)$.
  Answer: $x^{\tan x} \left( \frac{\tan x}{x} + \sec^2 x \ln x \right)$

III. Fundamental Concepts of Integration

• Problem: If $y = \int f(x)dx+C$, then $f(x)$ is called the -----
  Explanation: In the context of an integral $\int f(x)dx$, $f(x)$ is referred to as the integrand.
  Answer: integrand
• Problem: Which of the following equation is not correct?
  1. $\int_a^a f(x)dx = 0$ (Correct. The definite integral from a point to itself is zero.)
  2. $\int_a^b f(x)dx = \int_a^c f(x)dx$ (Incorrect. This statement implies that the integral over $[a,b]$ is the same as the integral over $[a,c]$, which is generally false unless $b=c$ or $f(x)=0$. The correct property is $\int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx$.)
  3. $|\int_a^b f(x)dx| \le \int_a^b |f(x)|dx$ if $a < b$ (Correct. This is the Triangle Inequality for Integrals.)
  4. $\int_a^b f(x)dx \le \int_a^b g(x)dx$ if $f(x) \le g(x)$ in $[a,b]$ (Correct. This is the Comparison Property of Integrals.)
  Answer: $\int_a^b f(x)dx = \int_a^c f(x)dx$