Title: CR_Int_17
Course: Demonstrating STACK (v4.3.7)

Miscellaneous integration exercises

Find \[ \int [value] d[value]. \]
Feedback: In order to find \[\int [value] d[value]\] we need to use the substitution \(u = [value]\). If \(u = [value]\) then \(du = [value] d[value]\), and so \(d[value] = \frac{du}[value]\). So, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] \frac{du}[value] \\ &\displaystyle =&\displaystyle \int [value] du \\ &\displaystyle =&\displaystyle [value] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Find \[ \int [value] d[value]. \]
Hint: let \(u = [value]\)
Feedback: In order to calculate \[\int [value] d[value]\] we can use the substitution \(u=[value]\). So, if \(u = [value]\) then \([value] = [value]\) and \(d[value] = [value] d[value]\), so, \(d[value] = d[value]/[value] = d[value]2([value])\). Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int \frac{1}{u} d[value]2([value]) \\ &\displaystyle =&\displaystyle \int [value]d[value] \\ &\displaystyle =&\displaystyle \int [value]d[value] \\ &\displaystyle =&\displaystyle [value] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Find \[ \int [value] d[value]. \]
Feedback: This problem can be solved using integration by parts. Define \(u = [value]\) and \(dv = [value]d[value]\). Then \(du = [value]d[value]\) and \(v = [value]\). So, \[\begin{array}{rcl}\displaystyle \int [value]d[value] &\displaystyle =&\displaystyle \int udv \\ &\displaystyle =&\displaystyle uv - \int vdu \\ &\displaystyle =&\displaystyle [value][value] - \int [value][value]d[value] \\ &\displaystyle =&\displaystyle [value] - \int [value]d[value] \\ &\displaystyle =&\displaystyle [value] - [value] \end{array}\]

Find \[ \int [value] d[value]. \]
Feedback: By linearity calc_int_linearity_rule: \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] d[value] + \int [value] d[value] + \int [value] d[value] \\ &\displaystyle =&\displaystyle [value]\int [value] d[value] + [value]\int [value] d[value] + \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] + [value] + [value] +c \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Find \[ \int [value] d[value]. \]
Feedback: By linearity calc_int_linearity_rule: \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] d[value] + \int [value] d[value] \\ &\displaystyle =&\displaystyle \int [value] d[value] + [value]\int [value] d[value] \\ &\displaystyle =&\displaystyle [value] + [value] +c \\ &\displaystyle =&\displaystyle [value] \end{array}\]

By taking logarithms to the base \(e\) establish that \(a^x\) can be written as \(e^{xlna}\). Hence find \[ \int [value] d[value]. \]
Feedback: Using the laws of logarithms \[e^{ln(x)} = x\] and \[ln(x^k) = kln(x)\] we get that \[\begin{array}{rcl}\displaystyle a^x &\displaystyle =&\displaystyle e^{ln(a^x)} \\ &\displaystyle =&\displaystyle e^{x\cdot ln(a)} \end{array}\] So, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int e^{x\cdot ln(a)} d[value] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Find \[ \int [value] d[value]. \]
Feedback: This problem can be solved using integration by parts [[hint:calc_int_methods_parts]]. Define \(u = [value]\) and \(dv = [value]d[value]\). Then \(du = [value]d[value]\) and \(v = [value]\). So, \[\begin{array}{rcl}\displaystyle \int [value]d[value] &\displaystyle =&\displaystyle \int udv \\ &\displaystyle =&\displaystyle uv - \int vdu \\ &\displaystyle =&\displaystyle [value][value] - \int [value][value]d[value] \\ &\displaystyle =&\displaystyle [value] - [value] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Find \[ \int [value] d[value]. \]
Feedback: In order to calculate \[\int [value] d[value]\] we need to express \[[value]\] as the sum of its partial fractions alg_partial_fractions. \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value] \\&\displaystyle =&\displaystyle [value][value] + [value][value] \end{array}\] By multiplying both sides by [value] we get \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value]+[value] \\&\displaystyle =&\displaystyle [value] \end{array}\] Equating coefficients of [value] we get a system of equations that can be solved to get \(A = 1\) and \(B = 1\). So, \[[value] = [value]\] Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle \int [value] d[value] + \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] +c \end{array}\]

Find \[ \int [value] d[value]. \]
Feedback: If \(u\) and \(v\) are both functions of \(x\) then \[\int u dv = uv - \int v du.\] Let \(u = [value]\) and \(dv = [value] d[value]\). Then \(du = [value] d[value]\) and \(v = [value]\). So, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int u dv \\&\displaystyle =&\displaystyle uv - \int v du \\&\displaystyle =&\displaystyle [value][value] - \int [value][value]d[value] +c \end{array}\] Hence, \[2\int [value] d[value] = [value]\] And so, \[\int [value] d[value] = \frac[value]{2}+ \frac{c}{2}.\]

Find \[ \int^[value]_[value] [value] d[value]. \]
Feedback: To solve this problem we will have to express [value] in a simpler way using some trigonometric identities trig_standard_identities Now, recall that if \(u\) and \(v\) are both functions of \(x\) then \[\int u dv = uv - \int v du.\] So, let \(u = [value]\) and \(dv = [value] d[value]\). Then \(du = [value] d[value]\) and \(v = [value]\). So, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int u dv \\ &\displaystyle =&\displaystyle uv - \int v du \\ &\displaystyle =&\displaystyle [[value]][[value]] - \int [[value]][[value]d[value]] +c \\ &\displaystyle =&\displaystyle [value] - \int [value]d[value] +c \\ &\displaystyle =&\displaystyle [value] + 2\int [value][[value]] d[value] +c \\ &\displaystyle =&\displaystyle [value] + 2\int [value] d[value] +c \end{array}\] Hence, \[3\int [value] d[value] = [value] + 2\int [value] d[value] +c \] And so, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle [value]+ [value] + \frac{c}{3} \\ &\displaystyle =&\displaystyle [value] + \frac{c}{3}. \end{array}\] Then \[\begin{array}{rcl}\displaystyle \int^[value]_[value] [value] d[value] &\displaystyle =&\displaystyle F([value]) - F([value]) \\ &\displaystyle =&\displaystyle [value] - \left[[value]\right] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Using the substitution \([value]=[value]\) find \[ \int [value] d[value]. \]
Feedback: We need to find \[\int [value] d[value]\] so, let \(x = [value]\). Then, \(dx = [value]du\). Hence, \[\begin{array}{rcl}\displaystyle \int \sqrt{1-[value]} d[value] & =&\displaystyle \int \sqrt{1-[value]}\, [value]du \\ & =&\displaystyle \int \sqrt[value]\, [value]du \\ & =&\displaystyle \int \cos(u)\cos(u)du \\ & =&\displaystyle \int \cos^2(u) du \\ & =&\displaystyle \int [value] du \\ & =&\displaystyle [value]+c \\ & =&\displaystyle [value]+c \end{array}\]

Find \[ \int [value] d[value]. \]
Feedback: In order to solve this problem we will have to use the method of integration by parts calc_int_methods_parts as follows: Let \(u = [value]\) and \(dv = [value] d[value]\). Then \(du = [value] d[value]\) and \(v = [value]\). So, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int u dv \\&\displaystyle =&\displaystyle uv - \int v du \\&\displaystyle =&\displaystyle [value][value] - \int [value][value]d[value] \end{array}\] In order to solve \[\int [value][value]d[value]\] we need to apply integration by parts again difining \(u = [value]\) and \(dv = [value] d[value]\). Then \(du = [value] d[value]\) and \(v = [value]\). So, \[\begin{array}{rcl}\displaystyle \int [value][value]d[value] &\displaystyle =&\displaystyle \int u dv \\&\displaystyle =&\displaystyle uv - \int v du \\&\displaystyle =&\displaystyle [value][value] - \int [value][[value]d[value]] \\&\displaystyle =&\displaystyle [value][value] - \int [value]d[value] \\&\displaystyle =&\displaystyle [value][value] + [value] \end{array}\] Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle [value][value] - \left([value][value] + [value]\right) \\&\displaystyle =&\displaystyle [value] \end{array}\] Thus, \[[value] = [value]\] And so, \[[value] = [value]\]

Find \[ \int [value] d[value]. \]
Feedback: In order to calculate \[\int [value] d[value]\] we need to express \[[value]\] as the sum of its partial fractions alg_partial_fractions. \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value] \\&\displaystyle =&\displaystyle [value][value] + [value][value] + [value][value] \end{array}\] By multiplying both sides by [value] we get \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value] + [value] + [value] \\&\displaystyle =&\displaystyle [value] \end{array}\] Equating coefficients of [value] we get a system of equations that can be solved to get \(A = [value]\), \(B = [value]\) and \(C = [value]\). So, \[[value] = [value]\] Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] d[value] \\&\displaystyle =&\displaystyle [value] \int [value] d[value] + [value] \int [value] d[value] + [value] \int [value] d[value] \\&\displaystyle =&\displaystyle [value] + [value] + [value] +c \\&\displaystyle =&\displaystyle [value] +c \end{array}\]

Find \[ \int [value] d[value]. \]
Feedback: Since the degree of the numerator is greater than the degree of the denominator we need to use long division to get a simpler form to express \[[value].\] alg_partial_fractions Using long division we get that \[[value] = [value] + [value]\] Now, \[[value] = [value][value] + [value][value]\] So, multiplying by [value] both sides we get \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value] + [value] \\&\displaystyle =&\displaystyle [value] \end{array}\] So, equating cefficients in [value] we get that a system of equations that can be solved to get \(A = [value]\) and \(B = [value]\) Hence, \[[value] = [value]\] So, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] d[value] \\&\displaystyle =&\displaystyle [value] \int [value] d[value] + [value] \int [value] d[value] + \int ([value]) d[value] \\&\displaystyle =&\displaystyle [value] + [value] + [value]+c \\&\displaystyle =&\displaystyle [value]+c \end{array}\]

Find \[ \int^[value]_[value] [value] d[value] \] by letting \([value]=2\sin(z)\).
Feedback: We want to solve \[\int [value] d[value]\] using the substitution \([value] = 2\sin(z)\). So, if \([value] = 2\sin(z)\) then \(d[value] = 2\cos(z)dz\) and \[\begin{array}{rcl}\displaystyle \int[value]d[value] &\displaystyle =&\displaystyle \int \frac{1}{[4-(2\sin(z))^2]^{3/2}}2\cos(z)dz \\ &\displaystyle =&\displaystyle \int \frac{2\cos(z)dz}{[4-4\sin^2(z)]^{3/2}} \\&\displaystyle =&\displaystyle \int \frac{2\cos(z)dz}{[4(1-\sin^2(z))]^{3/2}} \\ &\displaystyle =&\displaystyle \int \frac{2\cos(z)dz}{4^{3/2}[1-\sin^2(z)]^{3/2}} \\ &\displaystyle =&\displaystyle \int \frac{\cos(z)dz}{4[\cos^2(z)]^{3/2}} \\ &\displaystyle =&\displaystyle \int \frac{\cos(z)dz}{4\cos^3(z)} \\ &\displaystyle =&\displaystyle \int \frac{dz}{4\cos^2(z)} \\ &\displaystyle =&\displaystyle \frac{1}{4}\int \sec^2(z)dz \\ &\displaystyle =&\displaystyle \frac{\tan(z)}{4} +c \\ &\displaystyle =&\displaystyle \frac{1}{4}\tan\left[\sin^{-1}\left(\frac{t}{2}\right)\right]+c \end{array}\] We need \[\int^[value]_[value] [value] d[value]\] so, recall that if \[\int f(x) dx = F(x)+c\] then \[\int^a_b f(x) dx = F(a) - F(b).\] Thus, define \[F([value]) = [value]\] Then, \[\begin{array}{rcl}\displaystyle \int^[value]_[value] [value] d[value] &\displaystyle =&\displaystyle F([value]) - F([value]) \\ &\displaystyle =&\displaystyle [value] - [value] \\ &\displaystyle =&\displaystyle [value] \end{array}\]