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Word Problems on Integration

Samdom For Peace Prerequisite Topics:
Factoring
Partial Fractions
Exponents and Logarithms
Trigonometry
Differential Calculus

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions
Indicate the method(s) you used as applicable
Show all work
Simplify your answer(s)

(1.) Evaluate $\displaystyle\int -7dx$


Power Rule

$ \displaystyle\int -7dx \\[5ex] = \displaystyle\int -7 * 1 * dx \\[5ex] 1 = x^0 ...Law\:\:3...Exp \\[3ex] = \displaystyle\int -7x^0dx \\[5ex] = -7 \displaystyle\int x^0dx \\[5ex] = -7 * \dfrac{x^{0 + 1}}{0 + 1} + C \\[5ex] = -7 * \dfrac{x^1}{1} + C \\[5ex] = -7x + C $
(2.) Determine the antiderivative of $\cot x$


Integration by Algebraic Substitution

$ \displaystyle\int \cot xdx \\[5ex] \cot x = \dfrac{\cos x}{\sin x}...Quotient\:\:Identity \\[5ex] Let\:\: p = \sin x \\[3ex] \dfrac{dp}{dx} = \cos x \\[5ex] \dfrac{dx}{dp} = \dfrac{1}{\cos x} \\[5ex] dx = \dfrac{dp}{\cos x} \\[5ex] \rightarrow \\[3ex] \displaystyle\int \cot xdx = \displaystyle\int \dfrac{\cos x}{\sin x} dx \\[5ex] = \displaystyle\int \dfrac{\cos x}{p} * \dfrac{dp}{\cos x} \\[5ex] = \displaystyle\int \dfrac{dp}{p} \\[5ex] = \ln p + C \\[3ex] = \ln \sin x + C $
(3.) JAMB Evaluate $\displaystyle\int_1^3 (x^2 - 1)dx$

$ A.\:\: -6\dfrac{2}{3} \\[5ex] B.\:\: 6\dfrac{2}{3} \\[5ex] C.\:\: \dfrac{2}{3} \\[5ex] D.\:\: -\dfrac{2}{3} \\[5ex] $

Power Rule

$ \displaystyle\int_1^3 (x^2 - 1)dx \\[5ex] \displaystyle (x^2 - 1)dx = \displaystyle\int x^2dx - \displaystyle\int 1dx \\[5ex] 1 = x^0 ...Law\:\:3...Exp \\[3ex] = \dfrac{x^{2 + 1}}{2 + 1} - \displaystyle\int x^0dx \\[5ex] = \dfrac{x^3}{3} - \dfrac{x^{0 + 1}}{0 + 1} + C \\[5ex] = \dfrac{x^3}{3} - x + C \\[5ex] \rightarrow \\[3ex] \displaystyle\int_1^3 (x^2 - 1)dx = \left[\dfrac{x^3}{3} - x\right]_1^3 \\[5ex] = \left(\dfrac{3^3}{3} - 3\right) - \left(\dfrac{1^3}{3} - 1\right) \\[5ex] = (9 - 3) - \left(\dfrac{1}{3} - \dfrac{3}{3}\right) \\[5ex] = 6 - \left(-\dfrac{2}{3}\right) \\[5ex] = 6 + \dfrac{2}{3} \\[5ex] = 6\dfrac{2}{3} $
(4.) JAMB Evaluate $\displaystyle\int \sin 3xdx$

$ A.\:\: -\dfrac{1}{3}\cos 3x + c \\[5ex] B.\:\: \dfrac{1}{3}\cos 3x + c \\[5ex] C.\:\: \dfrac{2}{3}\cos 3x + c \\[5ex] C.\:\: -\dfrac{2}{3}\cos 3x + c \\[5ex] $

Integration by Algebraic Substitution

$ \displaystyle\int \sin 3xdx \\[5ex] Let\:\: p = 3x \\[3ex] \dfrac{dp}{dx} = 3 \\[5ex] \dfrac{dx}{dp} = \dfrac{1}{3} \\[5ex] dx = \dfrac{dp}{3} \\[5ex] \rightarrow \\[3ex] \displaystyle\int \sin 3xdx = \displaystyle\int \sin p * \dfrac{dp}{3} \\[5ex] = \dfrac{1}{3} * \displaystyle\int \sin pdp \\[5ex] = \dfrac{1}{3} * -\cos p + C \\[5ex] = \dfrac{1}{3} * - \cos 3x + C \\[3ex] = -\dfrac{1}{3}\cos 3x + c $
(5.) WASSCE:FM

(a) Express $\dfrac{x + 6}{(x + 1)^3}$ in partial fractions

(b) Use the answer in (a) to evaluate $\displaystyle\int_1^2 \dfrac{x + 6}{(x + 1)^3}dx$


$ (a.) \\[3ex] \dfrac{x + 6}{(x + 1)^3} \\[5ex] $ Numerator: cannot be simplified further
Denominator: cannot be simplified further
Form: Proper Fraction: Repeated Linear Factors at the Denominator

$ \dfrac{x + 6}{(x + 1)^3} \\[5ex] = \dfrac{A}{x + 1} + \dfrac{B}{(x + 1)^2} + \dfrac{C}{(x + 1)^3} \\[5ex] = \dfrac{A(x + 1)^2 + B(x + 1) + C}{(x + 1)^3} \\[5ex] = \dfrac{A[(x + 1)(x + 1)] + Bx + B + C}{(x + 1)^3} \\[5ex] = \dfrac{A[x^2 + x + x + 1] + Bx + B + C}{(x + 1)^3} \\[5ex] = \dfrac{A[x^2 + 2x + 1] + Bx + B + C}{(x + 1)^3} \\[5ex] = \dfrac{Ax^2 + 2Ax + A + Bx + B + C}{(x + 1)^3} \\[5ex] = \dfrac{Ax^2 + 2Ax + Bx + A + B + C}{(x + 1)^3} \\[5ex] $ Denominators are the same
Equate the numerators

$ x + 6 = Ax^2 + 2Ax + Bx + A + B + C \\[3ex] Swap \\[3ex] Ax^2 + 2Ax + Bx + A + B + C = x + 6 \\[3ex] Ax^2 + x(2A + B) + A + B + C = 0x^2 + x + 6 \\[3ex] \implies \\[3ex] A = 0...eqn.(1) \\[3ex] 2A + B = 1...eqn.(2) \\[3ex] A + B + C = 6...eqn.(3) \\[3ex] Substitute\;\;eqn.(1)\;\;in\;\;eqn.(2) \\[3ex] 2(0) + B = 1 \\[3ex] 0 + B = 1 \\[3ex] B = 1 - 0 \\[3ex] B = 1 \\[3ex] Substitute\;\;0\;\;for\;\;A\;\;and\;\;1\;\;for\;\;B\;\;in\;\;eqn.(3) \\[3ex] 0 + 1 + C = 6 \\[3ex] 1 + C = 6 \\[3ex] C = 6 - 1 \\[3ex] C = 5 \\[3ex] \implies \\[3ex] \dfrac{x + 6}{(x + 1)^3} = \dfrac{0}{x + 1} + \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] \dfrac{x + 6}{(x + 1)^3} = 0 + \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] \therefore \dfrac{x + 6}{(x + 1)^3} = \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] \underline{Check} \\[3ex] \dfrac{x + 6}{(x + 1)^3} = \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] \underline{RHS} \\[3ex] \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} \\[5ex] = \dfrac{1(x + 1) + 5}{(x + 1)^3} \\[5ex] = \dfrac{x + 1 + 5}{(x + 1)^3} \\[5ex] = \dfrac{x + 6}{(x + 1)^3} \\[5ex] = LHS \\[3ex] $ Integration by Partial Fractions

$ (b.) \\[3ex] \displaystyle\int_1^2 \dfrac{x + 6}{(x + 1)^3} dx \\[5ex] = \displaystyle\int_1^2 \left[\dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3}\right] dx \\[5ex] = \displaystyle\int_1^2 \dfrac{1}{(x + 1)^2} dx + \displaystyle\int_1^2 \dfrac{5}{(x + 1)^3} dx \\[5ex] Let\;\; p = x + 1 \\[3ex] \dfrac{dp}{dx} = 1 \\[5ex] \dfrac{dx}{dp} = \dfrac{1}{1} = 1 \\[5ex] dx = dp \\[3ex] \displaystyle\int \dfrac{1}{(x + 1)^2} dx \\[5ex] = \displaystyle\int \dfrac{1}{p^2} dp \\[5ex] = \displaystyle\int p^{-2} dp \\[5ex] = \dfrac{p^{-2 + 1}}{-2 + 1} \\[5ex] = \dfrac{p^{-1}}{-1} \\[5ex] = -p^{-1} \\[3ex] = -\dfrac{1}{p} \\[5ex] = -\dfrac{1}{x + 1} \\[5ex] \implies \\[3ex] \displaystyle\int_1^2 \dfrac{1}{(x + 1)^2} dx \\[5ex] = \left[-\dfrac{1}{x + 1}\right]_1^2 \\[5ex] = -\dfrac{1}{2 + 1} - -\dfrac{1}{1 + 1} \\[5ex] = -\dfrac{1}{3} + \dfrac{1}{2} \\[5ex] = \dfrac{-2 + 3}{6} \\[5ex] = \dfrac{1}{6} \\[7ex] \displaystyle\int \dfrac{5}{(x + 1)^3} dx \\[5ex] = 5\displaystyle\int \dfrac{1}{(x + 1)^3} dx \\[5ex] = 5\displaystyle\int \dfrac{1}{p^3} dp \\[5ex] = 5\displaystyle\int p^{-3} dp \\[5ex] = 5 * \dfrac{p^{-3 + 1}}{-3 + 1} \\[5ex] = 5 * \dfrac{p^{-2}}{-2} \\[5ex] = -\dfrac{5}{2} * p^{-2} \\[5ex] = -\dfrac{5}{2} * \dfrac{1}{p^2} \\[5ex] = -\dfrac{5}{2p^2} \\[5ex] = -\dfrac{5}{2(x + 1)^2} \\[5ex] \implies \\[3ex] \displaystyle\int_1^2 \dfrac{5}{(x + 1)^3} dx \\[5ex] = \left[-\dfrac{5}{2(x + 1)^2}\right]_1^2 \\[5ex] = -\dfrac{5}{2(2 + 1)^2} - -\dfrac{5}{2(1 + 1)^2} \\[5ex] = -\dfrac{5}{2(3)^2} + \dfrac{5}{2(2)^2} \\[5ex] = -\dfrac{5}{2(9)} + \dfrac{5}{2(4)} \\[5ex] = -\dfrac{5}{18} + \dfrac{5}{8} \\[5ex] = \dfrac{-20 + 45}{72} \\[5ex] = \dfrac{25}{72} \\[7ex] \implies \\[3ex] \displaystyle\int_1^2 \dfrac{1}{(x + 1)^2} dx + \displaystyle\int_1^2 \dfrac{5}{(x + 1)^3} dx \\[5ex] = \dfrac{1}{6} + \dfrac{25}{72} \\[5ex] = \dfrac{12 + 25}{72} \\[5ex] = \dfrac{37}{72} \\[5ex] \therefore \displaystyle\int_1^2 \dfrac{x + 6}{(x + 1)^3} dx = \dfrac{37}{72} $
(6.) KCSE A particle was moving along a straight line.
The acceleration of the particle after t seconds was given by $(4t - 13)\;ms^{-2}$.
The initial velocity of the particle was 18 $ms^{-1}$
(a) Determine the value of t when the particle is momentarily at rest.
(b) Find the distance covered by the particle between the time t = 1 second and t = 3 seconds.


$ a = 4t - 13 \\[3ex] a = \dfrac{dv}{dt} \\[5ex] dv = a dt \\[3ex] v = \displaystyle\int a dt \\[3ex] v = \displaystyle\int (4t - 13) dt \\[3ex] v = \dfrac{4t^2}{2} - 13t + C \\[3ex] v = 2t^2 - 13t + C \\[3ex] when\;\;t = 0,\;\;initial\;\;velocity = v = 18\;ms^{-1} \\[3ex] \implies \\[3ex] 18 = 2(0)^2 - 13(0) + C \\[3ex] 18 = 0 - 0 + C \\[3ex] 18 = C \\[3ex] C = 18 \\[3ex] v = 2t^2 - 13t + 18 \\[3ex] (a) \\[3ex] momentarily\;\;at\;\;rest\implies v = 0 \\[3ex] 0 = 2t^2 - 13t + 18 \\[3ex] 2t^2 - 13t + 18 = 0 \\[3ex] 2t(t - 2) - 9(t - 2) = 0 \\[3ex] (t - 2)(2t - 9) = 0 \\[3ex] t - 2 = 0 \;\;\;OR\;\;\; 2t - 9 = 0 \\[3ex] t = 2 \;\;\;OR\;\;\; 2t = 9 \\[3ex] t = 2\;s \;\;\;OR\;\;\; t = \dfrac{9}{2}\;s \\[3ex] (b) \\[3ex] v = \dfrac{dd}{dt} \\[5ex] dd = vdt \\[3ex] d = \displaystyle\int v dt \\[3ex] $ The zeros of the function...where the graph include areas above or below the x-axis (t-axis) are t = 2 seconds and t = $\dfrac{9}{2} = 4.5$ seconds
But we are asked to calculate the area (distance covered) between t = 1 second and t = 3 seconds
So, we have to first calculate the area between t = 1 second and t = 2 seconds
Then, we calculate the area between t = 2 seconds and t = 3 seconds
Because 4.5 seconds is greater than 3 seconds, we skip that area.

$ v = \displaystyle\int_1^2 (2t^2 - 13t + 18) dt + \displaystyle\int_2^3 (2t^2 - 13t + 18) dt \\[5ex] v = \left[\dfrac{2t^3}{3} - \dfrac{13t^2}{2} + 18t\right]_1^2 + \left[\dfrac{2t^3}{3} - \dfrac{13t^2}{2} + 18t\right]_2^3 \\[5ex] \left[\dfrac{2t^3}{3} - \dfrac{13t^2}{2} + 18t\right]_1^2 \\[5ex] t = 2\;s \\[3ex] \dfrac{2(2)^3}{3} - \dfrac{13(2)^2}{2} + 18(2) \\[5ex] \dfrac{16}{3} - \dfrac{26}{1} + \dfrac{36}{1} \\[5ex] \dfrac{16 - 78 + 108}{3} \\[5ex] \dfrac{46}{3}\;m \\[5ex] t = 1\;s \\[3ex] \dfrac{2(1)^3}{3} - \dfrac{13(1)^2}{2} + 18(1) \\[5ex] \dfrac{2}{3} - \dfrac{13}{2} + \dfrac{18}{1} \\[5ex] \dfrac{4 - 39 + 108}{6} \\[5ex] \dfrac{73}{6}\;m \\[5ex] Area = \dfrac{46}{3} - \dfrac{73}{6} = \dfrac{92 - 73}{6} = \dfrac{19}{6}\;m \\[5ex] \left[\dfrac{2t^3}{3} - \dfrac{13t^2}{2} + 18t\right]_2^3 \\[5ex] t = 3\;s \\[3ex] \dfrac{2(3)^3}{3} - \dfrac{13(3)^2}{2} + 18(3) \\[5ex] \dfrac{18}{1} - \dfrac{117}{2} + \dfrac{54}{1} \\[5ex] \dfrac{36 - 117 + 108}{2} \\[5ex] \dfrac{27}{2}\;m \\[5ex] t = 2\;s \\[3ex] \dfrac{46}{3}\;m \\[5ex] Area = \dfrac{27}{2} - \dfrac{46}{3} = \dfrac{81 - 92}{6} = -\dfrac{11}{6} \\[5ex] Because\;\;Area\;\;cannot\;\;be\;\;negative; \;\;Area = \dfrac{11}{6}\;m \\[5ex] Total\;\;Area = Distance\;\;covered \\[3ex] = \dfrac{19}{6} + \dfrac{11}{6} \\[5ex] = \dfrac{30}{6} \\[5ex] = 5\;m $
(7.) WASSCE:FM Evaluate $\displaystyle\int_0^1 \dfrac{3 - 3x^2}{x + 1}dx$

$ A.\:\: 1\dfrac{1}{3} \\[5ex] B.\:\: 1\dfrac{1}{2} \\[5ex] C.\:\: 3 \\[3ex] D.\:\: 4\dfrac{1}{2} $


Power Rule

$ \displaystyle\int_0^1 \dfrac{3 - 3x^2}{x + 1}dx \\[5ex] 3 - 3x^2 = 3(1 - x^2) \\[3ex] 1 - x^2 = 1^2 - x^2 = (1 + x)(1 - x)...Difference\:\:of\:\:Two\:\:Squares \\[3ex] 3 - 3x^2 = 3(1 + x)(1 - x) = 3(x + 1)(1 - x) \\[3ex] \rightarrow \\[3ex] \dfrac{3 - 3x^2}{x + 1} = \dfrac{3(x + 1)(1 - x)}{x + 1} = 3(1 - x) = 3 - 3x \\[5ex] \rightarrow \\[3ex] \displaystyle\int (3 - 3x)dx = \displaystyle\int 3dx - \displaystyle\int 3xdx \\[5ex] \displaystyle\int 3dx = 3x \\[5ex] \displaystyle\int 3xdx = 3 * \dfrac{x^{1 + 1}}{1 + 1} = \dfrac{3x^2}{2} \\[5ex] \rightarrow \\[3ex] \displaystyle\int_0^1 \dfrac{3 - 3x^2}{x + 1}dx = \left[3x - \dfrac{3x^2}{2}\right]_0^1 \\[5ex] = \left[3(1) - \dfrac{3(1)^2}{2}\right] - \left[3(0) - \dfrac{3(0)^2}{2}\right] \\[5ex] = \left[3 - \dfrac{3(1)}{2}\right] - \left[0 - \dfrac{3(0)}{2}\right] \\[5ex] = \left(3 - \dfrac{3}{2}\right) - \left(0 - \dfrac{0}{2}\right) \\[5ex] = \left(\dfrac{6}{2} - \dfrac{3}{2}\right) - (0 - 0) \\[5ex] = \dfrac{6 - 3}{2} - 0 \\[5ex] = \dfrac{3}{2} \\[5ex] = 1\dfrac{1}{2} $
(8.) Evaluate $\displaystyle\int (3^p + 7^p)dp$


Standard Integral

$ \displaystyle\int a^x dx = \dfrac{a^x}{\ln a} + C \\[5ex] \displaystyle\int (3^p + 7^p)dp = \displaystyle\int 3^pdp + \displaystyle\int 7^pdp \\[5ex] \displaystyle\int 3^p dx = \dfrac{3^p}{\ln 3} \\[5ex] \displaystyle\int 7^p dx = \dfrac{7^p}{\ln 7} \\[5ex] \rightarrow \\[3ex] \displaystyle\int (3^p + 7^p)dp = \dfrac{3^p}{\ln 3} + \dfrac{7^p}{\ln 7} + C $
(9.) Evaluate $\displaystyle\int \sqrt{2 + \sqrt{x}} dx$


Integration by Algebraic Substitution

$ \displaystyle\int \sqrt{2 + \sqrt{x}} dx \\[3ex] Let\:\: p = 2 + \sqrt{x} \\[3ex] \dfrac{dp}{dx} = \dfrac{1}{2\sqrt{x}} \\[5ex] \dfrac{dx}{dp} = 2\sqrt{x} \\[5ex] dx = 2\sqrt{x}dp \\[3ex] \rightarrow \\[3ex] \displaystyle\int \sqrt{2 + \sqrt{x}} dx = \displaystyle\int \sqrt{p} * 2\sqrt{x} dp \\[3ex] = 2\displaystyle\int \sqrt{p} * \sqrt{x} dp \\[3ex] p = 2 + \sqrt{x} \rightarrow \sqrt{x} = p - 2 \\[3ex] = 2 * \displaystyle\int \sqrt{p}(p - 2) dp \\[3ex] \displaystyle\int \sqrt{p}(p - 2) dp = \displaystyle\int p\sqrt{p}dp - \displaystyle\int 2\sqrt{p}dp \\[3ex] \displaystyle\int p\sqrt{p}dp = \displaystyle\int p^1 * p^{\dfrac{1}{2}}dp = \displaystyle\int p^{1 + \dfrac{1}{2}}dp = \displaystyle\int p^{\dfrac{3}{2}}dp \\[5ex] \displaystyle\int p\sqrt{p}dp = \dfrac{p^{\dfrac{3}{2} + 1}}{\dfrac{3}{2} + 1} \\[7ex] \displaystyle\int p\sqrt{p}dp = p^{\dfrac{5}{2}} \div \dfrac{5}{2} \\[5ex] \displaystyle\int p\sqrt{p}dp = p^{\dfrac{5}{2}} * \dfrac{2}{5} \\[5ex] p^{\dfrac{5}{2}} = (\sqrt{p})^5 = (\sqrt{p})^4 * \sqrt{p} = \left(p^{\dfrac{1}{2}}\right)^4 * \sqrt{p} = p^2\sqrt{p} \\[5ex] \displaystyle\int p\sqrt{p}dp = \dfrac{2p^2}{5} \\[5ex] \displaystyle\int 2\sqrt{p}dp = 2\displaystyle\int \sqrt{p}dp \\[3ex] \displaystyle\int \sqrt{p}dp = \dfrac{p^{\dfrac{1}{2} + 1}}{\dfrac{1}{2} + 1} \\[7ex] \displaystyle\int \sqrt{p}dp = p^{\dfrac{3}{2}} \div \dfrac{3}{2} \\[5ex] \displaystyle\int \sqrt{p}dp = p^{\dfrac{3}{2}} * \dfrac{2}{3} \\[5ex] p^{\dfrac{3}{2}} = (\sqrt{p})^3 = (\sqrt{p})^2 * \sqrt{p} = \left(p^{\dfrac{1}{2}}\right)^2 * \sqrt{p} = p\sqrt{p} \\[5ex] \displaystyle\int 2\sqrt{p}dp = \dfrac{2p\sqrt{p}}{3} \\[5ex] \rightarrow \\[3ex] displaystyle\int \sqrt{p}(p - 2) dp = \dfrac{2p^2\sqrt{p}}{5} - \dfrac{2p\sqrt{p}}{3} \\[5ex] = 2p\sqrt{p}\left(\dfrac{p}{5} - \dfrac{1}{3}\right) \\[5ex] = 2p\sqrt{p}\left(\dfrac{3p}{15} - \dfrac{5}{15}\right) \\[5ex] = 2p\sqrt{p}\left(\dfrac{3p - 5}{15}\right) \\[5ex] \rightarrow \\[3ex] 2 * \displaystyle\int \sqrt{p}(p - 2) dp = 2 * 2p\sqrt{p}\left(\dfrac{3p - 5}{15}\right) \\[5ex] = 4p\sqrt{p}\left(\dfrac{3p - 5}{15}\right) \\[5ex] Substitute\:\:back \\[3ex] = 4(2 + \sqrt{x})\sqrt{2 + \sqrt{x}}\left(\dfrac{3(2 + \sqrt{x}) - 5}{15}\right) \\[5ex] \dfrac{3(2 + \sqrt{x}) - 5}{15} = \dfrac{6 + 3\sqrt{x} - 5}{15} = \dfrac{1 + 3\sqrt{x}}{15} \\[5ex] \rightarrow \\[3ex] displaystyle\int \sqrt{2 + \sqrt{x}} dx = 4(2 + \sqrt{x})\left(\sqrt{2 + \sqrt{x}}\right)\left(\dfrac{1 + 3\sqrt{x}}{15}\right) \\[5ex] \therefore \displaystyle\int \sqrt{2 + \sqrt{x}} dx = \dfrac{4}{15}(2 + \sqrt{x})(1 + 3\sqrt{x})\left(\sqrt{2 + \sqrt{x}}\right) + C $
(10.) Find the integral of $\tan x$


Integration by Algebraic Substitution

$ \displaystyle\int \tan xdx \\[5ex] \tan x = \dfrac{\sin x}{\cos x}...Quotient\:\:Identity \\[5ex] Let\:\: p = \cos x \\[3ex] \dfrac{dp}{dx} = -\sin x \\[5ex] \dfrac{dx}{dp} = -\dfrac{1}{\sin x} \\[5ex] dx = -\dfrac{dp}{\sin x} \\[5ex] \rightarrow \\[3ex] \displaystyle\int \tan xdx = \displaystyle\int \dfrac{\sin x}{\cos x} dx \\[5ex] = \displaystyle\int \dfrac{\sin x}{p} * -\dfrac{dp}{\sin x} \\[5ex] = \displaystyle\int -\dfrac{dp}{p} \\[5ex] = -\ln p + C \\[3ex] = -\ln \cos x + C $
(11.)


(12.) Evaluate $\displaystyle\int (v^{-2} + 8v^{-1})dv$


Power Rule

$ \displaystyle\int (v^{-2} + 8v^{-1})dv \\[5ex] = \displaystyle\int v^{-2}dv + \displaystyle\int 8v^{-1}dv \\[5ex] = \dfrac{v^{-2 + 1}}{-2 + 1} + 8 * \displaystyle\int v^{-1}dv \\[5ex] = \dfrac{v^{-1}}{-1} + 8 * \ln v + C \\[5ex] = -v^{-1} + 8\ln v + C \\[5ex] = -\dfrac{1}{v} + 8\ln v + C \\[5ex] = 8\ln v - \dfrac{1}{v} + C $
(13.)


(14.)


(15.)


(16.)