I have a function:

a*tan²(pi/26*x) + 0.05

Whe I integrate this (from x=3 to x=2) I cannot find the right area.

Can someone help here?

First, use the fact that

tan²x = sec²x - 1

to rewrite your function as

a [sec²(pi/26 x) + 1] + 0.05

Put u = pi/26 x, so that du = pi/26 dx

Now, integrate, bearing in mind that the integral of sec²u is tan u.

Integral [a sec²(pi/26 x) + a + 0.05] dx

= 26/pi Integral [a sec²u + a + 0.05] du

= 26/pi [a tan u + (a + 0.05)u]

= 26/pi [a tan(pi/26 x) + (a + 0.05)(pi/26 x)]

Finally, evaluated between x=2 and x=3.

Hope this helps.

Thanks but did u make a mistake on the second line so it should have been:

a[sec²(π/26 x)-1] + 0.05

???????????

Yep, he did.
It happens to the best of us!