Evaluate : $\int$ $cos^4xdx\over {sin^3x{(sin^5x + cos^5x)^{3\over 5}}}$

Solution :

I = $\int$ $cos^4xdx\over {sin^3x{(sin^5x + cos^5x)^{3\over 5}}}$

= $\int$ $cos^4xdx\over {sin^6x{(1 + cot^5x)^{3\over 5}}}$ = $\int$ $cot^4xcosec^2xdx\over {{(1 + cot^5x)^{3\over 5}}}$

Put $1+cot^5x$ = t

$5cot^4xcosec^2x$dx = -dt

= -$1\over 5$ $\int$ $dt\over {t^{3/5}}$ = -$1\over 2$ $t^{2/5}$ + C

= -$1\over 2$ ${(1+cot^5x)}^{2/5}$ + C

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