Category: Week 4 group post (Page 2 of 3)

Group members: Shah A., Rafael D., Vishal R., David T., Anthony R.

Here are the solutions we have produced:

Anthony Regner

Here is Anthony’s solution for this problem (step-by-step):

1. Start by thinking what trig identity you should use. In this case, since it is the square root of (x² + a²) where a is a constant, you will use 1 = tan²θ = sec²θ.

2. Model your triangle where you put the square root of (x² + 9) as the hypotenuse, the number 3 as the adjacent leg, and the variable x as the opposite leg.

3. Let x = 3tanθ, and find its derivative. The dx = 3sec²θdθ. Then multiply values by dx/x.

4-5. Begin simplifying the values. In this case, 9 × 3 = 27, and then 27 ÷ 9 = 3.

6. Use the constant multiple rule by taking out the number 3 from the integral.

7. Factor out the square root of (9tan²θ + 1). It becomes secθ × 3.

8. Simplify: 3 × 1/3 = 1, and sec²θ divided by secθ equals to just secθ.

9. Take the trig functions in terms of cosine and sine. Then, simplify accordingly.

10. Perform u-substitution- since du/u fits the form perfectly. (Note in Step #10, it is (cosθ ÷ sin²θ) dθ.

11. Integrate the values by using the power rule.

12. After you integrate, add +C at the end, and simplify.

13.  Since -1 is in the denominator, multiply (or divide) the entire equation by negative 1.

14. Put the values back. In the event of using a negative exponent, use cscθ instead of sin^-1θ.

15. Put the trigonometric values in terms of x. Remember what you did to the model triangle. In case of cscθ, use the values of (hypotenuse / opposite).

The final answer is -sqrt(x² + 9) / x + C.

Group Members: Kevin V., Davide B., Kai H.-L., Jia Hong Y., Yanzu Z.