evaluate

Asked by  | 3rd Jan, 2010, 06:21: PM

Expert Answer:

sinx/(sinx+cosx) = 1/(1+cotx) = tanx/(1+tanx)

Put tanx = y

sec2x dx = dy

(1+tan2x)dx = dy

dx = dy/(1+tan2x)

dx = dy/(1+y2)

sinx dx/(sinx+cosx) = y dy/{(1+y)(1+y2)}

= (1/2){(y+1) dy/(1+y2)} - (1/2)dy/(1+y)

= (1/2)(1/2)ln(1+y2) + (1/2) tan-1y - (1/2)ln(1+y) + c

= (1/4)ln(1+y2) + (1/2) tan-1y - (1/2)ln(1+y) + c

= (1/4)ln(sec2x) + (x/2)  - (1/2)ln(1+tanx) + c

= ln((secx/(1+tanx) + x/2 + c.

Regards,

Team,

TopperLearning.

 

Answered by  | 3rd Jan, 2010, 10:04: PM

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