If int frac{d x}{sqrt{sin ^3 x cos x}}=g(x)+c | vedclass

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(B) Given, $I = \int \frac{d x}{\sqrt{\sin ^3 x \cos x}}$.We can rewrite the integrand as:$I = \int \frac{d x}{\sqrt{\sin ^4 x \cdot \frac{\cos x}{\si

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$\frac{-2}{\sqrt{	an x}}$

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(B) Given, $I = \int \frac{d x}{\sqrt{\sin ^3 x \cos x}}$.We can rewrite the integrand as:$I = \int \frac{d x}{\sqrt{\sin ^4 x \cdot \frac{\cos x}{\sin x}}} = \int \frac{d x}{\sin ^2 x \sqrt{\cot x}}$.Since $\frac{1}{\sin ^2 x} = \operatorname{cosec}^2 x$, we have:$I = \int \frac{\operatorname{cosec}^2 x}{\sqrt{\cot x}} d x$.Let $t = \cot x$. Then $dt = -\operatorname{cosec}^2 x d x$, which implies $\operatorname{cosec}^2 x d x = -dt$.Substituting these into the integral:$I = \int \frac{-dt}{\sqrt{t}} = -\int t^{-1/2} dt$.Integrating with respect to $t$:$I = -\frac{t^{1/2}}{1/2} + c = -2\sqrt{t} + c$.Substituting $t = \cot x$ back:$I = -2\sqrt{\cot x} + c = -\frac{2}{\sqrt{	an x}} + c$.Comparing this with $g(x) + c$, we get $g(x) = -\frac{2}{\sqrt{	an x}}$.

If $\int \frac{d x}{\sqrt{\sin ^3 x \cos x}}=g(x)+c$, then $g(x)$ is equal to

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