Integrate

1/(9+16cos^{2}x)=1/(9sin^{2}x+25cos^{2}x)=sec^{2}x/(9tan^{2}x+1)
put tanx=t & proceed....

Dividing num. and denominator by cos²x.
U will see sec²x appears in the denominator.
Now replace it by 1+ tan²x
Now put tan x = t
sec²x dx = dt
WE get \int_{0}^{\frac{\pi }{2}}{} \frac{dt}{9t^{2}+25}
we know \int \frac{dx}{a^{2}+ x^{2}}= \frac{1}{a}tan^{-1}\frac{x}{a}

Use it and get the result.

Hope u got it