InterviewSolution
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समाकलनो का मान ज्ञात कीजिएः `int_(pi//2)^(pi) (1-sin x)/(1-cos x)dx` |
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Answer» यहाँ `I=underset(pi//2)overset(pi)int (1-sin x)/(1-cos x)dx` `=underset(pi//2)overset(pi)int (1-sin""x/2cos""x/2)/(2sin^(2)""x/2)` `=underset(pi//2)overset(pi)int ((1)/(2)"cosec"^(2)pi/2-cot""x/2)dx` `=underset(pi//2)overset(pi)int 1/2"cosec"^(2) x/2 dx-underset(pi//2)overset(pi)int cot ""x/2dx` `=[1/2-(-cot""x/2)/(1/2)]_(pi//2)^(pi)-[(log|sin""x/2|)/(1/2)]_(pi//2)^(pi)` `=[cot""x/2]_(pi//2)^(pi)-[2log|sin""x/2|]_(pi//2)^(pi)` `=[-cot""pi/2+cot""pi/4]-[2 log sin ""pi/2-2 log sin ""pi/4]` `=(0+1)-[2log1-2log ""1/sqrt2]=1+2log""1/sqrt2` `=1-2log sqrt2=1-log 2` |
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