Menu Close

find-the-value-of-0-pi-4-cosx-sinx-tanx-dx-

Tinku Tara 0 Comments
Pin




Question Number 36057 by abdo mathsup 649 cc last updated on 28/May/18
find the value of ∫_0 ^(π/4) ((cosx)/(sinx +tanx))dx
findthevalueof∫0π4cosxsinx+tanxdx
Commented by abdo mathsup 649 cc last updated on 30/May/18
let I = ∫_0 ^(π/4) ((cosx)/(sinx +tanx))dx I = ∫_0 ^(π/4) ((cosx)/(sinx +((sinx)/(cosx))))dx = ∫_0 ^(π/4) ((cos^2 x)/(sinx.cosx +sinx))dx changement tan((x/2)) =t give I = ∫_0 ^((√2) −1) (((((1−t^2 )/(1+t^2 )))^2 )/(((2t (1−t^(2)) )/((1+t^2 )^2 )) +((2t)/(1+t^2 )))) ((2dt)/(1+t^2 )) = 2∫_0 ^((√2) −1) (((1−t^2 )^2 )/(2t(1−t^2 ) +2t(1+t^2 ))) (dt/(1+t^2 )) =2 ∫_0 ^((√2) −1) (((1−t^2 )^2 )/((1+t^2 )( 2t −2t^3 +2t +2t^3 )))dt = (1/2) ∫_0 ^((√2) −1) (((1−t^2 )^2 )/((1+t^2 )t)) dt ⇒ 2I = ∫_0 ^((√2)−1) ((t^4 −2t^2 +1)/(t^3 +t))dt =∫_0 ^((√2)−1) ((t(t^3 +t)−t^2 −2t^2 +1)/(t^(3 ) +t))dt = ∫_0 ^((√2)−1) t dt +∫_0 ^((√2) −1) ((−3t^2 +1)/(t^3 +t))dt but ∫_0 ^((√2) −1) tdt =[(t^2 /2)]_0 ^((√2)−1) =((((√2)−1)^2 )/2) let decompose F(t) = ((−3t^2 +1)/(t^3 +t)) F(t) = ((−3t^2 +1)/(t^3 +t)) = (a/t) +((bt +c)/(t^2 +1)) a=lim_(t→0) tF(t) = 1 =a lim_(t→+∞) t F(t) = −3 =a +b ⇒b=−3−a=−4 F(t) = (1/t) +((−4t +c)/(t^(2 ) +1)) F(1) =−1 = 1 +((−4+c)/2) =−1 +(c/2) ⇒c=0 ⇒ F(t)= (1/t) −((4t)/(t^2 +1)) ⇒ ∫_0 ^((√2) −1) ((−3t^2 +1)/(t^3 +t))dt = ∫_0 ^((√2)−1) { (1/t) −((4t)/(t^2 +1))}dt =[ln∣t∣ −2 ln(t^2 +1)]_0 ^((√2) −1) = [ln( (t/((t^2 +1)^2 )))]_0 ^((√2)−1) = ln((((√2) −1)/({((√2) −1)^2 +1}^( ))) −ln(0^+ ) =+∞ so this integral is divergent !
letI=∫0π4cosxsinx+tanxdxI=∫0π4cosxsinx+sinxcosxdx=∫0π4cos2xsinx.cosx+sinxdxchangementtan(x2)=tgiveI=∫02−1(1−t21+t2)22t(1−t2)(1+t2)2+2t1+t22dt1+t2=2∫02−1(1−t2)22t(1−t2)+2t(1+t2)dt1+t2=2∫02−1(1−t2)2(1+t2)(2t−2t3+2t+2t3)dt=12∫02−1(1−t2)2(1+t2)tdt⇒2I=∫02−1t4−2t2+1t3+tdt=∫02−1t(t3+t)−t2−2t2+1t3+tdt=∫02−1tdt+∫02−1−3t2+1t3+tdtbut∫02−1tdt=[t22]02−1=(2−1)22letdecomposeF(t)=−3t2+1t3+tF(t)=−3t2+1t3+t=at+bt+ct2+1a=limt→0tF(t)=1=alimt→+∞tF(t)=−3=a+b⇒b=−3−a=−4F(t)=1t+−4t+ct2+1F(1)=−1=1+−4+c2=−1+c2⇒c=0⇒F(t)=1t−4tt2+1⇒∫02−1−3t2+1t3+tdt=∫02−1{1t−4tt2+1}dt=[ln∣t∣−2ln(t2+1)]02−1=[ln(t(t2+1)2)]02−1=ln(2−1{(2−1)2+1}()−ln(0+)=+∞sothisintegralisdivergent!
Answered by tanmay.chaudhury50@gmail.com last updated on 28/May/18
∫_0 ^(Π/4) ((cos^2 x)/(sinxcosx+sinx )) dx (−1)∫_0 ^(Π/4) ((1−cos^2 x−1)/(sinx(1+cosx)))dx (−1)∫_0 ^(Π/4) ((1−cosx)/(sinx))dx+∫_0 ^(Π/4) ((sinx)/(sin^2 x(1+cosx)))dx ∫_0 ^(Π/4) ((cotx−cosecx)/)−∫_1 ^(1/( (√2))) (dt/((1−t^2 )(1+t))) ∣lnsinx−lntan(x/2)∣_0 ^(Π/4) +I_2 I_2 =∫_1 ^(1/( (√2))) (A/(1+t))+(B/(1−t))+(C/((1+t)^2 )) dt (1/((1+t)^2 (1−t)))=(A/(1+t))+(B/(1−t))+(C/((1+t)^2 )) 1=A(1+t)(1−t)+B(1+t)^2 +C(1−t) 1=A(1−t^2 )+B(1+2t+t^2 )+C(1−t) A−At^2 +B+2Bt+Bt^2 +C−Ct=1 A+B+C+t(2B−C)+t^2 (B−A)=1 A+B+C=1 2B−C=0 B−A=0 A=B 2B−C=0 C=2B A+B+C=1 B+B+2B=1 B=(1/4) A=(1/4) C=(1/2) ∫_1 ^(1/( (√2))) ((1/4)/(1+t))dt+∫_1 ^(1/( (√2))) ((1/4)/(1−t))dt+∫_1 ^(1/( (√2))) ((1/2)/((1+t)^2 ))dt ∣(1/4)ln(1+t)−(1/4)ln(1−t)−(1/2)×(1/(1+t))∣_1 ^(1/( (√2)))
∫0Π4cos2xsinxcosx+sinxdx(−1)∫0Π41−cos2x−1sinx(1+cosx)dx(−1)∫0Π41−cosxsinxdx+∫0Π4sinxsin2x(1+cosx)dx∫0Π4cotx−cosecx−∫112dt(1−t2)(1+t)∣lnsinx−lntanx2∣0Π4+I2I2=∫112A1+t+B1−t+C(1+t)2dt1(1+t)2(1−t)=A1+t+B1−t+C(1+t)21=A(1+t)(1−t)+B(1+t)2+C(1−t)1=A(1−t2)+B(1+2t+t2)+C(1−t)A−At2+B+2Bt+Bt2+C−Ct=1A+B+C+t(2B−C)+t2(B−A)=1A+B+C=12B−C=0B−A=0A=B2B−C=0C=2BA+B+C=1B+B+2B=1B=14A=14C=12∫112141+tdt+∫112141−tdt+∫11212(1+t)2dt∣14ln(1+t)−14ln(1−t)−12×11+t∣112
Answered by sma3l2996 last updated on 28/May/18
A=∫_0 ^(π/4) ((cosx)/(sinx+tanx))dx=∫_0 ^(π/4) ((cos^2 x)/(cosxsinx+sinx))dx =∫_0 ^(π/4) ((1−sin^2 x)/(sinx(cosx+1)))dx=∫_0 ^(π/4) (dx/(sinx(cosx+1)))−∫_0 ^(π/4) ((sinx)/(1+cosx))dx =∫_0 ^(π/4) (dx/(sinx(cosx+1)))+[ln∣1+cosx∣]_0 ^(π/4) let u=tanx/2⇒dx=((2du)/(1+u^2 )) sinx=2cos(x/2)sin(x/2)=((2u)/(1+u^2 )) 1+cosx=2cos^2 (x/2)=(2/(1+u^2 )) A=ln((((√2)+1)/2))−ln2+∫_0 ^(tan^(−1) (π/8)) (1/(((2u)/(1+u^2 ))×(2/(1+u^2 ))))×((2du)/(1+u^2 )) A=ln((√2)+2)−2ln2+(1/2)∫_0 ^((√2)−1) ((1+u^2 )/u)du =ln((√2)+2)−2ln2+(1/2)[lnu+(1/2)u^2 ]_0 ^((√2)−1) A=−∞ because (lim_(u→0) lnu=−∞)
A=∫0π/4cosxsinx+tanxdx=∫0π/4cos2xcosxsinx+sinxdx=∫0π/41−sin2xsinx(cosx+1)dx=∫0π/4dxsinx(cosx+1)−∫0π/4sinx1+cosxdx=∫0π/4dxsinx(cosx+1)+[ln∣1+cosx∣]0π/4letu=tanx/2⇒dx=2du1+u2sinx=2cos(x/2)sin(x/2)=2u1+u21+cosx=2cos2(x/2)=21+u2A=ln(2+12)−ln2+∫0tan−1(π/8)12u1+u2×21+u2×2du1+u2A=ln(2+2)−2ln2+12∫02−11+u2udu=ln(2+2)−2ln2+12[lnu+12u2]02−1A=−∞because(limlnuu→0=−∞)

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *