int frac{sin^3 x + cos^3 x}{sin^2 x cos^2 x} | vedclass

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(D) We are given the integral $I = \int \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} \, dx$. Dividing each term in the numerator by the denominator,w

Vedclass · Accepted Answer

$\sec x - \csc x + c$

Vedclass · Answer

(D) We are given the integral $I = \int \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} \, dx$. Dividing each term in the numerator by the denominator,we get: $I = \int \left( \frac{\sin^3 x}{\sin^2 x \cos^2 x} + \frac{\cos^3 x}{\sin^2 x \cos^2 x} ight) \, dx$ $I = \int \left( \frac{\sin x}{\cos^2 x} + \frac{\cos x}{\sin^2 x} ight) \, dx$ Using trigonometric identities $\frac{\sin x}{\cos^2 x} = \sec x 	an x$ and $\frac{\cos x}{\sin^2 x} = \csc x \cot x$,we have: $I = \int (\sec x 	an x + \csc x \cot x) \, dx$ Integrating term by term: $\int \sec x 	an x \, dx = \sec x$ and $\int \csc x \cot x \, dx = -\csc x$. Therefore,$I = \sec x - \csc x + c$.

$\int \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} \, dx = $

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