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9 months ago

```							differentiate both sides we getf(x)sinxcosx= f’(x)/[2(b^2 – a^2)f(x)] where we have used chain rule to diff lnf(x).let y= f(x) so thatdy/dx= 2(b^2 – a^2)*y^2sinxcosxor dy/y^2= (b^2 – a^2)*2sinxcosx dxintegrate both sides – 1/y= 2(b^2 – a^2)*sin^2x + cor y= 1/[(a^2 – b^2)*sin^2x – c] where c is any constant.if we let c= – b^2, theny= 1/[(a^2sin^2x + b^2(1 – sin^2x)]or y= f(x)= 1/(a^2sin^2x + b^2cos^2x)KINDLY APPROVE :))
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9 months ago
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