InterviewSolution
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InterviewSolution
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Evaluate : (i) `int((4-5 cosx)/(sin^(2)x))dx` , (ii) `int((1-cos2x)/(1+cos2x)) dx` (iii) `int (1)/(sin^(2)x cos^(2)x) dx` , (iv) `int(cos2x)/(cos^(2)x sin^(2) x) dx` |
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Answer» (i) `int((4-5cosx)/(sin^(2)x))dx = int(4/(sin^(2)x) - (5cosx)/(sin^(2)x))dx` `= int(4cosec^(2) x - 5 cosec x cot x) dx` `= 4 intcosec^(2) xdx - 5 int cosec x cot x dx` `= 4 (-cotx) - 5(-cosecx) + C` `= - 4 cot x + 5 cosec x + C`. (ii) `int((1-cos 2x)/(1+ cos2x)) dx = int(2sin^(2)x)/(2 cos^(2)x) dx = inttan^(2)x dx` `= int(sec^(2)x - 1)dx = intsec^(2)x dx - intdx` `= tanx-x + C`. (iii) `int (1)/(sin^(2)x cos^(2)x) dx = int((sin^(2)x +cos^(2)x)/(sin^(2)xcos^(2)x))dx` `= int(1/(cos^(2)x) + 1/(sin^(2)x)) dx` `= intsec^(2) xdx + intcosec^(2)xdx = tan x - cot x +C`. (iv) `int(cos2x)/(cos^(2)x sin^(2)x) dx = int((cos^(2)x - sin^(2)x)/(cos^(2)x sin^(2)x)) dx` `= int((1)/(sin^(2)x) - (1)/(cos^(2)x))dx` `= intcosec^(2)x dx - int sec^(2)x dx = -cotx - tan x+ C`. |
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