InterviewSolution
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InterviewSolution
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`int(sinx)/sqrt(1+sinx)dx` का मान ज्ञात कीजिए। |
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Answer» दिया है -` I = int(sinx)/(sqrt(1+sin x))dx = int((1+sin x) - 1)/(sqrt(1+sin x))` ` = int sqrt(1+sinx)dx - int(dx)/(sqrt(1+sin x))` `= intsqrt(cos^(2).(x)/(2)+sin^(2)+(x)/(2)+2sin.(x)/(2)cos.(x)/(2)dx)` `-int(dx)/(sqrt(cos^(2)(x//2)+sin^(2)(x//2)+2sin(x//2))cos(x//2))` `= int[cos((x)/(2))+sin((x)/(2))]dx - int(dx)/([cos (x//2)+sin(x//2)])` ` = (2sin.(x)/(2) - cos .(x)/(2)) -(1)/(2).int(dx)/((1)/(sqrt(2)cos.(x)/(2)+(1)/(sqrt(2))sin.(x)/(2)))` `=(2sin.(x)/(2)-2cos.(x)/(2)) - (1)/(sqrt(2)) . int(dx)/(sin((pi)/(2) +(pi)/(2)))` ` = (2sin.(x)/(2) - 2cos.(x)/(2)) -(1)/(sqrt(2)). intcosec((x)/(2) +(pi)/(4))dx` ` = 2 (sin.(x)/(2)-cos.(x)/(2)) - (1)/(sqrt(2)) xx 2 log[ tan ((x)/(4) +(pi)/(8))]+c` ` = 2(sin.(x)/(2) - cos.(x)/(2))- sqrt(2)[ tan ((pi)/(4) +(pi)/8)]+c` |
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