If int frac{1}{a^2 sin^2 x + b^2 cos^2 x} dx | vedclass

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(A) We have the integral $I = \int \frac{dx}{a^2 \sin^2 x + b^2 \cos^2 x}$.Dividing numerator and denominator by $\cos^2 x$,we get $I = \int \frac{\se

Vedclass · Accepted Answer

$\sqrt{40}$

Vedclass · Answer

(A) We have the integral $I = \int \frac{dx}{a^2 \sin^2 x + b^2 \cos^2 x}$.Dividing numerator and denominator by $\cos^2 x$,we get $I = \int \frac{\sec^2 x dx}{a^2 	an^2 x + b^2}$.Let $u = 	an x$,then $du = \sec^2 x dx$.The integral becomes $I = \int \frac{du}{a^2 u^2 + b^2} = \frac{1}{a^2} \int \frac{du}{u^2 + (b/a)^2}$.Using the formula $\int \frac{dx}{x^2 + k^2} = \frac{1}{k} 	an^{-1}(\frac{x}{k}) + C$,we get $I = \frac{1}{a^2} \cdot \frac{a}{b} 	an^{-1}(\frac{u}{b/a}) + C = \frac{1}{ab} 	an^{-1}(\frac{a}{b} 	an x) + C$.Comparing this with the given expression $\frac{1}{12} 	an^{-1}(3 	an x) + C$,we have $ab = 12$ and $\frac{a}{b} = 3$.From $\frac{a}{b} = 3$,we get $a = 3b$. Substituting into $ab = 12$,we get $(3b)b = 12 \implies 3b^2 = 12 \implies b^2 = 4 \implies b = 2$ (assuming $a, b > 0$).Then $a = 3(2) = 6$.The expression is $6 \sin x + 2 \cos x$. The maximum value of $A \sin x + B \cos x$ is $\sqrt{A^2 + B^2}$.Maximum value = $\sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40}$.

List-$I$	List-$II$
$1. \int \frac{\sin^2 x}{\cos^4 x} dx$	$A. \frac{\tan^2 x}{2} + \ln\|\cos x\| + c$
$2. \int \frac{\sin^4 x}{\cos^2 x} dx$	$B. \cos x + \sec x + c$
$3. \int \frac{\sin^3 x}{\cos^2 x} dx$	$C. \frac{\tan^3 x}{3} + c$
$4. \int \frac{\sin^3 x}{\cos^3 x} dx$	$D. \tan x + \frac{\sin 2x}{4} - \frac{3x}{2} + c$
	$E. \cos x - \sec x + c$

If $\int \frac{1}{a^2 \sin^2 x + b^2 \cos^2 x} dx = \frac{1}{12} \tan^{-1}(3 \tan x) + C$,then the maximum value of $a \sin x + b \cos x$ is:

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