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फलन `int(cosx)/(cos 3x)` का मान ज्ञात कीजिए। |
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Answer» `int(cosx)/(cos 3x)dx=int(cosx)/(4cos^(3)x-3cosx)dx=int(1)/(4cos^(2)x-3)dx` `=int(1)/(4cos^(2)x-3(cos^(2)x+sin^(2)x))dx` `=int(1)/(cos^(2)x-3sin^(2)x)dx=int(1)/(cos^(2)x(1-3tan^(2)x))dx` `=int(sec^(2)x)/(1-3tan^(2)x)dx=(1)/(sqrt3)int(dt)/(1-t^(2))` ( यहाँ `sqrt3 tanx =t rArr sec^(2)x dx=(1)/(sqrt3)dt`) `=(1)/(sqrt3).(1)/(2.1)log((1+t)/(1-t))+c` `=(1)/(2sqrt3)log((1+sqrt3 tanx)/(1-sqrt3 tanx))+c` `=(1)/(2sqrt3)log((1+sqrt3tanx)/(1-sqrt3tanx))+c` |
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