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Question Number 159870 by tounghoungko last updated on 21/Nov/21

∫ ((1−cot^2 x)/(1+sin x)) dx =?

∫1−cot2x1+sinxdx=?

Answered by Ar Brandon last updated on 21/Nov/21

I=∫((1−cot^2 x)/(1+sinx))dx=∫(((1−cot^2 x)(1−sinx))/(1−sin^2 x))dx =∫((1−sinx−cot^2 x+cot^2 xsinx)/(cos^2 x))dx =∫((1/(cos^2 x))−((sinx)/(cos^2 x))−(1/(sin^2 x))+(1/(sinx)))dx =tanx−(1/(cosx))+cotx+ln∣cosecx−cotx∣+C =((sinx−1)/(cosx))+((cosx)/(sinx))+ln∣((1−cosx)/(sinx))∣+C

I=∫1−cot2x1+sinxdx=∫(1−cot2x)(1−sinx)1−sin2xdx=∫1−sinx−cot2x+cot2xsinxcos2xdx=∫(1cos2x−sinxcos2x−1sin2x+1sinx)dx=tanx−1cosx+cotx+ln∣cosecx−cotx∣+C=sinx−1cosx+cosxsinx+ln∣1−cosxsinx∣+C

Commented by tounghoungko last updated on 22/Nov/21

yes...very simple

yes...verysimple

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