4This pretty simpler
Thanks :)
If ∫ [ 0 to pi/2 ] dx / [a2 cos2 x + b2 sin2x ] = pi/2ab
Then the value of
∫[0 to pi/2] dx / [ 4 cos2x + 9 sin2x ]2 = ???
(A) 11 pi / 864
(B) 13 pi/864
(C) 17 pi/864
(D) None
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5 Answers
4Thanks Tapas
But the formula is tooooo tough to remember
(NO OTHER WAY TO SOLVEEEEEEEEEEEE ???????????????????????))
21\int_{0}^{\pi/2}{\frac{dx}{(a^2cos^2x+b^2sin^2x)}}=\frac{\pi}{2ab}\\
differntiate the given eq firstly wrt to a....i
and then wrt to b....ii
add them i.e i and ii and then see wat happens.
u will get the resulting integral as \frac{\pi(a^2+b^2)}{4a^3b^3}