Title: CR_Int_16
Course: Demonstrating STACK (v4.3.7)

Integration using trigonometrical identities

Find \[ \int [value] d[value]. \]
Feedback: \[ \int [value] d[value] = [value]+c\]

Using the identity \[\sin^2(A)=[value]\] to find \[ \int [value] d[value]. \]
Feedback: Since \[[value] = [value]\] it follows, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle [value]+c \end{array}\]

Use the identity \[\sin(A)\cos(A)=\frac{\sin(2A)}{2}\] to find \[ \int [value] d[value]. \]
Feedback: Since \[[value] = [value]\] it follows that \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle [value]+c \end{array}\]

Find \[ \int [value] d[value]. \]
Feedback: For this question we will have to use the identity \[\sin(A)\cos(B) = \frac{1}{2} [\sin(A+B) + \sin(A-B)]\] as follows: \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value]\frac{1}{2} [\sin([value] + [value]) + \sin([value] - [value]) \\ &\displaystyle =&\displaystyle [value][value] \\ &\displaystyle =&\displaystyle [value] \end{array}\] So, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Find \[ \int [value] d[value]. \] Hint: Use a trigonometric identity and note that the derivative of \(\tan(t)\) is \(\sec^2(t)\)
Feedback: For this question we need to use the identity \[tan^2(A) = sec^2(A) - 1\] as follows: \[[value] = [value]\] So, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle \int [value] d[value] + \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Find \[ \int [value] d[value]. \] Hint: square an identity for \(\cos^2(A)\).
Feedback: For this question we need to use the identity \[cos^2(A) = \frac{1}{2}[1+cos(2A)]\] as follows: \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [[value]]^2\\ &\displaystyle =&\displaystyle \left[[value]\right]^2 \\ &\displaystyle =&\displaystyle [value] \end{array}\] So, \[ \int [value] d[value] = \int [value] d[value] \] To solve this integral we need to express \[[value]\] in a simpler form using the same identity: \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value][value] \\ &\displaystyle =&\displaystyle [value]\left[[value]\right] \\ &\displaystyle =&\displaystyle [value] \end{array}\] Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] + [value] + [value] d[value] \\ &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] \end{array}\]