If e c^2 x ( ec ,x , + , ,x ) ^ 9 2 ,dx = ( ec ,x , - , ,x )^ 7 2… - Vidyadip

MCQ

If $\int {\frac{{\cos e{c^2}x}}{{{{\left( {\cos ec\,x\, + \,\cot \,x} \right)}^{\frac{9}{2}}}}}\,dx} $ = ${\left( {\cos ec\,x\, - \,\cot \,x} \right)^{\frac{7}{2}}}\left( {\frac{1}{\alpha } + \frac{{{{\left( {\cos ec\,x\, - \,\cot \,x} \right)}^2}}}{{11}}} \right) + \,C$ (where $C$ is constant of integration and $\alpha \in N)$ , then $\alpha $ is

A
$5$
B
$\frac {7}{2}$
C
$10$
✓
$7$

✓

Answer

Correct option: D.

$7$

d
$I = \int {\frac{{\cos e{c^2}x}}{{{{\left( {\cos ecx + \cot x} \right)}^{9/2}}}}} dx$

Put $\cos ecx + \cot x = z$

$\cos ec - \cot x = \frac{1}{z}.$

$ - 2\cos e{c^2}xdx = \left( {1 + \frac{1}{{{z^2}}}} \right)dz$

$\therefore I = - \frac{1}{2}\int {\frac{{1 + \frac{1}{{{z^2}}}}}{{{z^{9/2}}}}} dz$

$ = \frac{1}{2}\left[ {\int {{z^{ - 9/2}}dz + \int {{z^{\frac{{ - 13}}{2}}}} dz} } \right]$

$ = - \frac{1}{2}\left[ {\frac{{{z^{ - 7/2}}}}{{\left( { - 7} \right)}}2 + \frac{{{z^{ - 11/2}}}}{{\left( { - 11} \right)}}2} \right] + C$

$ = {z^{\frac{{ - 7}}{2}}}\left[ {\frac{1}{7} + \frac{{{z^{ - 3}}}}{{11}}} \right] + C$

$ = {\left( {\cos ecx - \cot x} \right)^{\frac{7}{2}}}\left( {\frac{1}{7} + \frac{{{{\left( {\cos ecx - \cot x} \right)}^2}}}{{11}}} \right) + C$

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