Menu Close

I-sec-x-1-csc-x-dx-

Tinku Tara 0 Comments
Pin




Question Number 141847 by iloveisrael last updated on 24/May/21
I = ∫ ((sec x)/(1+csc x)) dx
I=∫secx1+cscxdx
Answered by MJS_new last updated on 24/May/21
∫((sec x)/(1+csc x))= [t=tan (x/2) → dx=2cos^2 (x/2) dt] =−4∫(t/((t−1)(t+1)^3 ))dt= =−(1/2)∫(dt/(t−1))−2∫(dt/((t+1)^3 ))+∫(dt/((t+1)^2 ))+(1/2)∫(dt/(t+1))= =−(1/2)ln (t−1) +(1/((t+1)^2 ))−(1/(t+1))+(1/2)ln (t+1) = =−(t/((t+1)^2 ))+(1/2)ln ((t+1)/(t−1)) =... =−((sin x)/(2(1+sin x)))+(1/2)ln ∣((cos x)/(1−sin x))∣ +C
∫secx1+cscx=[t=tanx2→dx=2cos2x2dt]=−4∫t(t−1)(t+1)3dt==−12∫dtt−1−2∫dt(t+1)3+∫dt(t+1)2+12∫dtt+1==−12ln(t−1)+1(t+1)2−1t+1+12ln(t+1)==−t(t+1)2+12lnt+1t−1=…=−sinx2(1+sinx)+12ln∣cosx1−sinx∣+C
Answered by iloveisrael last updated on 24/May/21
I= ∫ ((1/(cos x))/(1+(1/(sin x)))) dx = ∫ ((sin x)/(cos xsin x+cos x)) dx I= ∫ ((sin x)/(cos x(sin x+1))) dx I=∫ ((sin x(sin x−1))/(−cos^3 x)) dx I=−∫ ((sin^2 x−sin x)/(cos^3 x)) dx I=−{∫ sec^3 x dx −∫ sec x dx }− ∫ ((d(cos x))/(cos^3 x)) I=ln ∣sec x+tan x∣+ (1/2)sec^2 x − ∫ sec^3 x dx
I=∫1cosx1+1sinxdx=∫sinxcosxsinx+cosxdxI=∫sinxcosx(sinx+1)dxI=∫sinx(sinx−1)−cos3xdxI=−∫sin2x−sinxcos3xdxI=−{∫sec3xdx−∫secxdx}−∫d(cosx)cos3xI=ln∣secx+tanx∣+12sec2x−∫sec3xdx

Pin

Leave a Reply

Your email address will not be published.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