Title: Integration #1
Course: Demonstrating STACK (v4.4.7)

Evaluate the following integral, using a substitution if necessary: \[ \int [value] d[value] \]
Feedback: Using a substitution, let \[u=[value]-[value][value]\] Differentiating \(u\) with respect to [value] gives \[du=-[value]d[value]\] Therefore \[\begin{array}{rcl}\displaystyle \int [value]d[value] &\displaystyle = &\displaystyle \int-\frac{1}[value]\ e^u du\\ &\displaystyle =&\displaystyle -\frac{1}[value]\ e^u +c\\ &\displaystyle =&\displaystyle -\frac{1}[value][value]+c\end{array}\]

Complete the square in the quadratic expression q which is given by \[ q= [value], \] and use this to evaluate the following integral \[ \int [value]d[value] \]
Feedback: Completing the square for \(q\) gives \[q=[value][value] + [value][value] + [value]=[value][value]+ [value][value]+[value]+[value]=([value][value]+[value])^2 + [value]\] Therefore \[\int[value]d[value] = \int\frac{1}[value]d[value]=\int\frac{1}[value]d[value]\] Let \([value]+[value] = \sinh(u)\) giving \[d[value] = [value]du\] Therefore \[\int[value]d[value] = \int\frac{1}[value][value]du\] and since \(\cosh^2(u)-\sinh^2(u) = 1\) \[\begin{array}{rcl}\displaystyle \int\frac{1}[value][value]du & \displaystyle =&\displaystyle\int\frac{1}[value][value]du\\ \int \frac{du}[value] & \displaystyle =& \displaystyle \frac{u}[value] + c \\ & \displaystyle =&\displaystyle [value] \end{array}\]

Evaluate the following integral, using a substitution if necessary: \[ \int \sinh([value][value]-[value])d[value] \]
Feedback: Using a substitution, let \[u=[value][value]-[value]\] Differentiating \(u\) with respect to [value] gives \[du=[value]d[value]\] Therefore \[ \int \sinh([value][value]-[value])\ d[value] =\int \frac{1}[value]\sinh(u)du\] The integral of sinh is cosh giving \[ \int \frac{1}[value]\sinh(u)du = \frac{1}[value]\cosh([value][value]-[value]) + c\]

Evaluate the following integral, using a substitution if necessary: \[ \int [value][value]^[value]([value]^[value]-[value])^\frac[value][value] d[value] \]
Feedback: Using a substitution, let \[u=[value]^[value]-[value]\] Differentiating \(u\) with respect to [value] gives \[du=[value][value]^[value]d[value]\] Therefore \[\begin{array}{rcl}\displaystyle \int [value][value]^[value]([value]^[value]-[value])^\frac[value][value]\ d[value] &\displaystyle = &\displaystyle\int u^[value] du \\ &\displaystyle = &\displaystyle [value]u^[value]+c \\ &\displaystyle = &\displaystyle [value]([value]^[value]-[value])^[value]+c \end{array}\]