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