how will u integrate 1/[4(sin x)^2 + 5(cos x)^2] w.r.t x?

Question

plz give sol.

Answer 1

solution is

sqrt(5)arctan sqrt(81- tanx) all over 100

first you simplify everything so you can get to a form where you can use the equation

du/(a+bcos u)

which should equal

2/(sqrt(a^2 + b^2) x [arctan (a^2 - b^(2) tan (u/2)] / [a+b]

there is a value C, but no value is given in your equation

Answer 2

1/[4(sin x)^2 + 5(cos x)^2]
=1/[4(sin x)^2 + 4(cos x)^2 + (cos x)^2]
=1/[4 + (cos x)^2]
=1/[2^2 + (cos x)^2]
...

Answer 3

the solution is

the solution for integral 1/[a^2(sin x)^2 + b^2(cos x)^2] is
1/(ab)arctg(a*tgx/b)+C

in ur case it will be
1/(2*sqrt(5))*arctg(2*tg(x)/sqrt(5))

yeah and almost forgot +C :D

Answer 4

FOR THIS KIND OF PROBLEM..YOU MUST USE THE INTEGRATION TECHNIQUE CALLED "PARTIAL FRACTION"!!!

SORRY DONT HAVE TIME TO GIVE YOU THE FULL ANSWER! THE TECHNIQUE IS IN GOOGLE!