If int {frac{{csc^2 x}}{{{{left( {csc x + cot | vedclass

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Question

(D) Let $I = \int {\frac{{\csc^2 x}}{{{{\left( {\csc x + \cot x} \right)}^{9/2}}}}} dx$. Let $z = \csc x + \cot x$. Then $\frac{dz}{dx} = -\csc x \cot

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(D) Let $I = \int {\frac{{\csc^2 x}}{{{{\left( {\csc x + \cot x} \right)}^{9/2}}}}} dx$. Let $z = \csc x + \cot x$. Then $\frac{dz}{dx} = -\csc x \cot x - \csc^2 x = -\csc x(\cot x + \csc x) = -\csc x \cdot z$. Also,$\csc x - \cot x = \frac{1}{z}$. Adding the two equations: $2\csc x = z + \frac{1}{z} \implies \csc x = \frac{z^2 + 1}{2z}$. Substituting into the derivative: $dz = -\left( \frac{z^2 + 1}{2z} \right) z dx = -\frac{z^2 + 1}{2} dx$. Thus,$dx = -\frac{2}{z^2 + 1} dz$. Substituting into $I$: $I = \int \frac{(\csc x)^2}{z^{9/2}} \left( -\frac{2}{z^2 + 1} \right) dz = \int \frac{(\frac{z^2+1}{2z})^2}{z^{9/2}} \left( -\frac{2}{z^2 + 1} \right) dz$. $I = -\frac{1}{2} \int \frac{z^2+1}{z^{13/2}} dz = -\frac{1}{2} \int (z^{-9/2} + z^{-13/2}) dz$. $I = -\frac{1}{2} \left[ \frac{z^{-7/2}}{-7/2} + \frac{z^{-11/2}}{-11/2} \right] + C = \frac{z^{-7/2}}{7} + \frac{z^{-11/2}}{11} + C$. Since $z^{-1} = \csc x - \cot x$,we have $I = (\csc x - \cot x)^{7/2} \left( \frac{1}{7} + \frac{(\csc x - \cot x)^2}{11} \right) + C$. Comparing this with the given expression,we find $\alpha = 7$.

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