Let : x = (3/5) sin² F.

Then : dx = (3/5) 2 sin F cos F dF.

Also : v(3-5x) = v3. cos F.
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From (1), then,

I = ? [ ( v3. cos F ) / ( 2(3/5) sin² F ) ] · (3/5) 2 sin F cos F dF

= v3. ? ( cos² F / sin F ) dF

= v3. ? [ ( 1 - sin² F ) / sin F ] dF

= v3. ? ( csc F - sin F ) dF

= v3. { ln | csc F - cot F | - ( - cos F ) } + C

= v3. { ln | csc F - cot F | + cos F } + C

= v3. ln | (v(5x/3) - v(3-5x)/v(5x) | + v(3-5x) + C .......... Ans.
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