InTeGrAl

Take cos²x common from the bracket term in the denominator ... and also from numerator .... cancel the cosine terms.

The expr. stands ... ∫(5tan x + 6)(2sin ²x + 3 cos²x)√(tan²x + 4)dx

Then take cos²x common from the other term in denominator and put it in numerator as sec²x.

The expr. now is: ∫(5tan x + 6)sec²x(2tan²x + 3)√(tan²x + 4)dx

Take tan x = z
sec²x dx = dz

The expr. stands:

∫(5z+6) dz(2z²+3)√(z²+4)

Then break the expr. into:

5∫z dz(2z²+3)√(z²+4) + 6∫dz(2z²+3)√(z²+4)

Now the expr. can be easily evaluated ....

I just figured out that the 2nd part is going to give u a little bit of problem but the first part is easy as hell!!!

In first part take (z²+4) = t²

∫t dt(2t²-5) - 6/5 * ∫2(z²+4) - (2z²+3) (2z²+3)√(z²+4)dz

now I'm sure u can do the first part.

The only trouble lies in the second part ....

2nd part: -12/5 * ∫√(z²+4)
(2z²+3)dz + 6/5 * ∫dz√(z²+4)dz

Now The expr. is: -12/5 * ∫√(z²+4)(2z²+3)dz + 3/5 * tan^-1 z/2

Please give me some time .... I will try to come up with a simplified solution to the 2nd part.

ok..............take ur time...............

I just figured out the second part ... it could be done in a better way.

2nd part:

6∫dzz³(2+3z^-2)√(1+4z^-2)

Now take 1+4z^-2 = t²

The expr. now is: -6/4 * ∫t dt(3t²+5)4t

=-6∫dt(3t²+5)

Now I'm am sure that u have enough knowledge and intellect to calculate this part ..... hope there are no calculation mistakes...