int_{0}^{pi /2}{frac{dx}{{{a}^{2}}{{cos }^{2} | vedclass

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(D) Let $I = \int_0^{\pi /2} {\frac{{dx}}{{{a^2}{{\cos }^2}x + {b^2}{{\sin }^2}x}}.} $Dividing the numerator and denominator by ${\cos ^2}x$,we get$I

Vedclass · Accepted Answer

$\frac{\pi }{{2ab}}$

Vedclass · Answer

(D) Let $I = \int_0^{\pi /2} {\frac{{dx}}{{{a^2}{{\cos }^2}x + {b^2}{{\sin }^2}x}}.} $Dividing the numerator and denominator by ${\cos ^2}x$,we get$I = \int_0^{\pi /2} {\frac{{\sec^2 x \, dx}}{{{a^2} + {b^2}	an^2 x}}} $Substituting $b 	an x = t$,we have $b \sec^2 x \, dx = dt$,or $\sec^2 x \, dx = \frac{dt}{b}$.When $x = 0$,$t = 0$. When $x = \frac{\pi}{2}$,$t 	o \infty$.Therefore,$I = \int_0^\infty {\frac{{dt/b}}{{{a^2} + {t^2}}}} = \frac{1}{b} \int_0^\infty \frac{dt}{a^2 + t^2}$Using the formula $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$,we get$I = \frac{1}{b} \left[ \frac{1}{a} 	an^{-1} \left( \frac{t}{a} ight) ight]_0^\infty = \frac{1}{ab} \left( 	an^{-1}(\infty) - 	an^{-1}(0) ight)$$I = \frac{1}{ab} \left( \frac{\pi}{2} - 0 ight) = \frac{\pi}{2ab}$.

$\int_{0}^{\pi /2}{\frac{dx}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}}\,=$

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