Solveee

If âˆ« [ 0 to pi/2 ] dx / [a² cos² x + b² sin²x ] = pi/2ab

Then the value of

âˆ«[0 to pi/2] dx / [ 4 cos²x + 9 sin²x ]² = ???

(A) 11 pi / 864

(B) 13 pi/864

(D) None

5 Answers

Thanks Tapas

But the formula is tooooo tough to remember

(NO OTHER WAY TO SOLVEEEEEEEEEEEE ???????????????????????))

\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}

This pretty simpler

Thanks :)

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