The value of the integral int frac{dx}{sin^2 | vedclass

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(B) Let $I = \int \frac{dx}{\sin^2 x + 	an^2 x}$.Since $\sin^2 x = \frac{	an^2 x}{1 + 	an^2 x}$, we have $I = \int \frac{dx}{\frac{	an^2 x}{1 +

Vedclass · Accepted Answer

$-\frac{1}{2 	an x} - \frac{1}{2\sqrt{2}} 	an^{-1} \left(\frac{	an x}{\sqrt{2}}ight) + c$

Vedclass · Answer

(B) Let $I = \int \frac{dx}{\sin^2 x + 	an^2 x}$.Since $\sin^2 x = \frac{	an^2 x}{1 + 	an^2 x}$, we have $I = \int \frac{dx}{\frac{	an^2 x}{1 + 	an^2 x} + 	an^2 x} = \int \frac{(1 + 	an^2 x) dx}{	an^2 x + 	an^2 x(1 + 	an^2 x)} = \int \frac{\sec^2 x \ dx}{	an^2 x + 	an^2 x + 	an^4 x} = \int \frac{\sec^2 x \ dx}{2	an^2 x + 	an^4 x}$.Let $u = 	an x$, then $du = \sec^2 x \ dx$. The integral becomes $I = \int \frac{du}{u^4 + 2u^2} = \int \frac{du}{u^2(u^2 + 2)}$.Using partial fractions: $\frac{1}{u^2(u^2 + 2)} = \frac{A}{u^2} + \frac{B}{u^2 + 2}$. Solving gives $A = \frac{1}{2}$ and $B = -\frac{1}{2}$.So, $I = \frac{1}{2} \int \frac{du}{u^2} - \frac{1}{2} \int \frac{du}{u^2 + 2} = \frac{1}{2} (-\frac{1}{u}) - \frac{1}{2} \cdot \frac{1}{\sqrt{2}} 	an^{-1} \left(\frac{u}{\sqrt{2}}ight) + c$.Substituting $u = 	an x$, we get $I = -\frac{1}{2 	an x} - \frac{1}{2\sqrt{2}} 	an^{-1} \left(\frac{	an x}{\sqrt{2}}ight) + c$.

The value of the integral $\int \frac{dx}{\sin^2 x + \tan^2 x}$ is...

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