The value of int frac{sec x cdot tan x}{9-16 | vedclass

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(B) Let $I = \int \frac{\sec x 	an x}{9-16 	an ^2 x} \,d x$. Using the identity $	an^2 x = \sec^2 x - 1$,we get: $I = \int \frac{\sec x 	an x}{9-1

Vedclass · Accepted Answer

$\frac{1}{40} \log \left(\frac{5+4 \sec x}{5-4 \sec x}\right)+c$,(where $c$ is a constant of integration)

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(B) Let $I = \int \frac{\sec x 	an x}{9-16 	an ^2 x} \,d x$. Using the identity $	an^2 x = \sec^2 x - 1$,we get: $I = \int \frac{\sec x 	an x}{9-16(\sec^2 x - 1)} \,d x = \int \frac{\sec x 	an x}{25-16 \sec^2 x} \,d x$. Let $\sec x = t$,then $\sec x 	an x \,d x = dt$. Substituting these into the integral: $I = \int \frac{dt}{5^2 - (4t)^2}$. Using the standard integral formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} ight| + c$,where $x$ is replaced by $4t$ and we divide by the derivative of $4t$ (which is $4$): $I = \frac{1}{4} \cdot \frac{1}{2(5)} \log \left| \frac{5+4t}{5-4t} ight| + c = \frac{1}{40} \log \left| \frac{5+4 \sec x}{5-4 \sec x} ight| + c$.

The value of $\int \frac{\sec x \cdot \tan x}{9-16 \tan ^2 x} \,d x$ is equal to

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