Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 85776 by M±th+et£s last updated on 24/Mar/20

Answered by MJS last updated on 24/Mar/20

∫((a sin^2  x +2b sin x cos x +c cos^2  x)/(sin x +cos x))dx=       [t=tan (x/2) → dx=(2/(t^2 +1))dt]  =−2∫((ct^4 −4bt^3 +2(2a−c)t^2 +4bt+c)/((t^2 +1)^2 (t−1−(√2))(t−1+(√2))))dt=  =2(a+2b−c)∫((t−((a−2b−c)/(a+2b−c)))/((t^2 +1)^2 ))dt+       +(a−2b−c)∫(dt/(t^2 +1))−       −(((√2)(a−2b+c))/4)∫(dt/(t−1−(√2)))+       +(((√2)(a−2b+c))/4)∫(dt/(t−1+(√2)))  I think you can solve these

asin2x+2bsinxcosx+ccos2xsinx+cosxdx=[t=tanx2dx=2t2+1dt]=2ct44bt3+2(2ac)t2+4bt+c(t2+1)2(t12)(t1+2)dt==2(a+2bc)ta2bca+2bc(t2+1)2dt++(a2bc)dtt2+12(a2b+c)4dtt12++2(a2b+c)4dtt1+2Ithinkyoucansolvethese

Commented by M±th+et£s last updated on 24/Mar/20

yes sir thank you its easy to solve now

yessirthankyouitseasytosolvenow

Commented by MJS last updated on 24/Mar/20

you can always use this substitution  (“Weierstrass”) for integrals with  trigonometric functions  t=tan (x/2) ⇔ x=2arctan t  → dx=2cos^2  (x/2) dt=(1+cos x)dt=(2/(t^2 +1))dt  sin x =((2t)/(t^2 +1))  cos x =−((t^2 −1)/(t^2 +1))  the main problem is to factorize the  denominator...

youcanalwaysusethissubstitution(Weierstrass)forintegralswithtrigonometricfunctionst=tanx2x=2arctantdx=2cos2x2dt=(1+cosx)dt=2t2+1dtsinx=2tt2+1cosx=t21t2+1themainproblemistofactorizethedenominator...

Commented by MJS last updated on 24/Mar/20

...also with sinh x, cosh x:  t=tanh (x/2) ⇔ x=2arctanh t  → dx=2cosh^2  (x/2) dt=(1+cosh x)dt=−(2/(t^2 −1))dt  sinh x =−((2t)/(t^2 −1))  cosh x =−((t^2 +1)/(t^2 −1))

...alsowithsinhx,coshx:t=tanhx2x=2arctanhtdx=2cosh2x2dt=(1+coshx)dt=2t21dtsinhx=2tt21coshx=t2+1t21

Terms of Service

Privacy Policy

Contact: info@tinkutara.com