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`int_(0)^(pi//2)x sin x dx ` का मान ज्ञात कीजिए ।

Answer» `int_(0)^(pi//2)x sin x dx=[-x cos x]_(0)^(pi//2)-int_(0)^(pi//2)1.(-cos x)dx`
`" "` (खण्डशः समाकलन करने पर )
`=[-x cos x]_(0)^(pi//2)+int_(0)^(pi//2)cos xdx`
`=[-x cos x]_(0)^(pi//2)+[sinx]_(0)^(pi//2)`
`=[-(pi)/(2)cos.(pi)/(2)+0cos0]+[sin.(pi)/(2)-sin0]`
`=(-(pi)/(2)xx0+0xx1)+(-10)=1`


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