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समाकलनो का मान ज्ञात कीजिएः `int_(pi//2)^(pi) (1-sin x)/(1-cos x)dx`

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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