Question

Find the function f(x) that has the given derivative and the given value.

f'(x)=sin(x)+2sec(x)tan(x) and f(pi)=1

Answer 1

Integrate this with respect to x. The sin(x) part is straightforward. The other, I will check the table of derivatives, but I am pretty sure it's a u substitution. You will get F(x) + C. Then evaluate F(pi) and choose an appropriate constant.

Answer 2

take the anti derivative, and then ... what bmp said
not a u-sub though

Answer 3

u = secx, du = tanxsecxdx? Then it's the integral of 1du.

Answer 4

anti derivative of sine is \(-\cos(x)\) and anti derivative of \(2\sec(x)\tan(x)\) is \(2\sec(x)\)

Answer 5

I think you can do it like that, at least. We are left with \[2\int\limits 1du\]

Answer 6

i guess so but it is the same as saying \(u=-\cos(x), du=\sin(x)dx\) and and so for the first one you have
\[\int1du \]

Answer 7

in other words you know the derivative of \(\sec(x)\) is \(\sec(x)\tan(x)\) so you know the anti derivative of \(\sec(x)\tan(x)\) is \(\sec(x)\)

Answer 8

That was a bit dumb of my part, haha. That makes perfect sense.

Answer 9

in any case you know the anti derivative is
\[f(x)=-\cos(x)+2\sec(x)+C\] and you know
\[f(\pi)=-\cos(\pi)+2\sec(\pi)+C\] so you can solve for C

Answer 10

no not dumb, just another way of looking at it