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

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

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

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(B) Given,$\int \frac{d x}{\sqrt{\sin ^3 x \cos x}}=g(x)+c$.We can rewrite the integral as:$\int \frac{d x}{\sqrt{\sin ^4 x \cdot \frac{\cos x}{\sin x}}} = \int \frac{d x}{\sin ^2 x \sqrt{\cot x}}$This simplifies to:$\int \operatorname{cosec}^2 x \cdot (\cot x)^{-1/2} d x$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:$\int -t^{-1/2} dt = -\frac{t^{1/2}}{1/2} + c = -2\sqrt{t} + c$Substituting back $t = \cot x$:$-2\sqrt{\cot x} + c = -\frac{2}{\sqrt{	an x}} + c$Therefore,$g(x) = -\frac{2}{\sqrt{	an x}}$.

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

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