Given f(x) = the integral from 0 to x?

Question

f(x) = the integral from 0 to x

(t^2-49)/(1+cos^2t) dt

... at what value of x does the local max of f(x) occur?

What I have is... FTC1 lets me get rid of integration sign so all I have left is:

t^2-49)/(1+cos^2t

find the 2nd derivative to find the critical points and set it to 0.

I'm having trouble dealing with the trig functions what I have so far from the quotient rule is

f"(x)=0=((1+cos^2t)(2t)) - ((t^2-49)(-2costsint))

but then ok - when i figure out t - that will be.... what exactly? An inflection point right? But I'm looking for a maximum value. Hmmm... Ideas?

Answer 1

Your approach is not quite right if I understand the question correctly. The Second (not the first) Fundamental Theorem of Calculus tells us that f'(x)=(x^2-49)/(1+cos^2(x))

Now we can just set that equal to 0 and see that it equals 0 at x= -7 and x=7, and then you can use the first derivative test with a factor line analysis to see which is a local maximum.

No quotient rule involved. It's a lot simpler than you think.

Answer 2

FTC1 says what you have left is:
dy/dx = (x^2-49)/(1+cos^2x)
If you set this = 0 you get x= +/- 7 [cos^x<> -1].
There are other maximum values but not as large as the ones occuring at x=7 and x = -7.