I'm having a hard time understanding how to do Integrals involving tan^2. I have two problems:
1. Find the integral of (tan^2 y +1)dy
2. Find the integral of (7tan^2 u +15)du
1. My approach to it is to replace the tan^2 y portion of the problem with sec^2 y -1, but it doesn't give me the answer that was given once I've worked it out. Can someone explain to me what I'm doing wrong and how I need to approach these problems?
2. For the latter question, I have no idea how they got the solution 7tanu +8u+C. None as far as tan is concerned and even less of an idea of how they got 8u.
don't forget your trig identities
tan^2+1 = sec^2
∫sec^2 y dy
is easy, right?
7tan^2 u + 15 = 7tan^2 u+7 + 8
= 7sec^2 u + 8
Now we're back to
∫7sec^2 u + 8 du
Thank you very much for your help Steve. I see now where I was going wrong with the problems and feel slightly embarrassed. Thanks again! :)
To solve the first problem, let's break it down step by step:
1. Start with the integral ∫(tan^2 y + 1)dy.
2. Rewrite tan^2 y as sec^2 y - 1, since tan^2 y = sec^2 y - 1.
3. Now we have the integral ∫(sec^2 y - 1 + 1)dy, which simplifies to ∫sec^2 ydy.
4. Recall that the integral of sec^2 ydy is equivalent to the derivative of tan y. So the integral becomes ∫sec^2 ydy = tan y.
5. Finally, we have the solution to the integral ∫(tan^2 y + 1)dy = tan y + C, where C is the constant of integration.
Therefore, the correct answer to the first problem is tan y + C.
Now let's move on to the second problem:
1. Start with the integral ∫(7tan^2 u + 15)du.
2. Since we don't have a direct formula for the integral of tan^2 u, we need to manipulate the equation to simplify it.
3. Notice that 7tan^2 u + 15 can be factored as 7(tan^2 u + 15/7).
4. The term (tan^2 u + 15/7) can be rewritten as (tan^2 u + 1 - 1 + 15/7), where we separated 15/7 into 1 and 15/7.
5. Simplify further to get (tan^2 u + 1) + (-1 + 15/7).
6. The first portion, (tan^2 u + 1), can be solved using the approach we discussed earlier, which gives us tan u.
7. The second portion, (-1 + 15/7), simplifies to -2/7.
8. Therefore, the integral of (7tan^2 u + 15)du becomes the integral of (tan^2 u + 1)du + (-2/7)du.
9. Using our previous result, we know that the integral of (tan^2 u + 1)du is tan u.
10. The integral of (-2/7)du is (-2/7)u.
11. Combining these results, the final solution is tan u - (2/7)u + C, where C is the constant of integration.
So, the correct answer to the second problem is 7tan u - (2/7)u + C, or you can also write it as 7tan u + (-2/7)u + C, whichever format you prefer.
I hope this explanation helps you understand how to approach and solve integrals involving tan^2.