int left( frac{1}{x^2} + frac{sin^3 x + cos^3 | vedclass

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Question

(B) We have the integral $I = \int \left( \frac{1}{x^2} + \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} \right) dx$. Split the integral into two parts

Vedclass · Accepted Answer

$-\frac{1}{x} + \sec x + \csc x + c$

Vedclass · Answer

(B) We have the integral $I = \int \left( \frac{1}{x^2} + \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} ight) dx$. Split the integral into two parts: $I = \int \frac{1}{x^2} dx + \int \frac{\sin^3 x}{\sin^2 x \cos^2 x} dx + \int \frac{\cos^3 x}{\sin^2 x \cos^2 x} dx$. Simplify each term: $I = \int x^{-2} dx + \int \frac{\sin x}{\cos^2 x} dx + \int \frac{\cos x}{\sin^2 x} dx$. Evaluate the integrals: $\int x^{-2} dx = -\frac{1}{x}$. For the second term,let $u = \cos x$,then $du = -\sin x dx$,so $\int \sec x 	an x dx = \sec x$. For the third term,let $v = \sin x$,then $dv = \cos x dx$,so $\int \csc x \cot x dx = -\csc x$. Combining these,we get $I = -\frac{1}{x} + \sec x - \csc x + c$. This can be written as $-\frac{1}{x} + \frac{1}{\cos x} - \frac{1}{\sin x} + c = -\frac{1}{x} + \frac{\sin x - \cos x}{\sin x \cos x} + c$.

$\int \left( \frac{1}{x^2} + \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} \right) dx =$

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