Anthonyk2013
Junior Member
- Joined
- Sep 15, 2013
- Messages
- 132
There are certainly other ways to solve these integrals, but I would solve these two by converting the trig functions to their sine and cosine components. For 9a:
\(\displaystyle 5\int cot\left(2t\right)\cdot csc\left(2t\right)dt=5\int \frac{cos\left(2t\right)}{sin\left(2t\right)}\cdot\frac{1}{sin\left(2t\right)}dt=5\int \frac{cos\left(2t\right)}{sin^2\left(2t\right)}dt\)
Use u-substitution, where u = sin(2t). Then for 9b:
\(\displaystyle \frac{4}{3}\int sec\left(4t\right)\cdot tan\left(4t\right)dt=\frac{4}{3}\int \frac{1}{cos\left(4t\right)}\cdot \frac{sin\left(4t\right)}{cos\left(4t\right)}dt= \frac{4}{3}\int \frac{sin\left(4t\right)}{cos^2\left(4t\right)}dt\)
Again, use u-substitution here. What would you choose as your u?
In (9)(a), you've got a cotangent and a cosecant. Do you see any integral formula (from your table) that includes both of these? Maybe try using that formula. Use the same method for (9)(b).Confused as to what rules I use in these questions.
In (9)(a), you've got a cotangent and a cosecant. Do you see any integral formula (from your table) that includes both of these? Maybe try using that formula. Use the same method for (9)(b).
If you get stuck, please reply showing your thoughts and efforts so far. Thank you!![]()