How do you use partial fractions to find the integral int (sec^2x)/(tanx(tanx+1))dx∫sec2xtanx(tanx+1)dx?

1 Answer
Andrea S.
Jan 30, 2017

int sec^2x/(tanx(tanx+1))dx = ln abs (sinx/(sinx+cosx)) +C∫sec2xtanx(tanx+1)dx=ln∣∣∣sinxsinx+cosx∣∣∣+C

Explanation:

Substitute:

t = tanxt=tanx

dt = sec^2x dxdt=sec2xdx

so we have:

int sec^2x/(tanx(tanx+1))dx = int (dt)/(t(t+1))∫sec2xtanx(tanx+1)dx=∫dtt(t+1)

Now we can solve the integral using partial fractions:

1/(t(t+1)) = A/t+B/(t+1)1t(t+1)=At+Bt+1

1/(t(t+1)) = (A(t+1)+ Bt)/(t(t+1))1t(t+1)=A(t+1)+Btt(t+1)

1= (A+B)t +A1=(A+B)t+A

{(A = 1),(B=-1):}

int (dt)/(t(t+1)) = int (1/t-1/(t+1))dt

int (dt)/(t(t+1)) = int (dt)/t-int (dt)/(t+1)

int (dt)/(t(t+1)) = ln abs(t) -ln abs(t+1) +C =ln abs (t/(t+1)) +C

and substituting back x:

int sec^2x/(tanx(tanx+1))dx = ln abs (tanx/(tanx+1)) +C

int sec^2x/(tanx(tanx+1))dx = ln abs (sinx/(sinx+cosx)) +C

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