OK I am actually going to show u what I did here with this function to get the area. I can tell that the result is wrong though coz I know the area under the curve. Please tell me where I went wrong:

f(X) = d [Tan² {(a - x) b}] + c

= d [Tan² (ba - bx)] + c

let u = ba - bx

dx = 1/-b du

d/-b ∫ Tan² (u) du + ∫ c dx

= d/-b [Tan u - u] + cx

= d/-b [Tan (ba - bx) - (ba - bx)] + cx

Anything wrong???

[font=times new roman][size=3]
d/-b ∫ Tan² (u) du + ∫ c dx

= d/-b [Tan u - u] + cx

Anything wrong???
i don't think the integral of tan²x is tan x

I didn't write that! Look again.

Anyway becasue nobody had responded for so long I presumed it was right and had a look again. It turns out that the integration was right but when I did the calculation I put a bracket in the wrong place :rolleyes: so its ok.

Thanks for helping by not responding :D

I didn't write that! Look again.

Anyway becasue nobody had responded for so long I presumed it was right and had a look again. It turns out that the integration was right but when I did the calculation I put a bracket in the wrong place :rolleyes: so its ok.

Thanks for helping by not responding :D
oops... my fault. yeah, it seems OK then.