Grade 12Integral Calculus
Integral of (sinx-cosx)/(1+sinxcosx) with limits zero to pi/2
3 Answers
Chandan
Divide numerator and denominator by cos2x.
=∫sec2x/(sec2x+tanx)dx.
put tanx=t.
sec2x dx=dt.
also
sec2x=1+tan2x.
=∫dt/(t2+1+t).
=add and subtrac1/4 from the denominator.
=∫dt/(t2-1/4+1/4+t+1).
=∫dt/((t-1/2)2+1/4).
=
((t-1/2)/1/2) ∫da/(a2+k2)=(a/k)
=(2t-1)+c
=((2tanx-1))+c
Arjun Radhakrishnan
change x=(
– x)
then you will get
add both of them(i.e I+I =2I)
2*I = 
2*I=0
I = 0
therefore the answer is 0
Kushagra Madhukar
Dear student,
Please find the solution to your problem.
We have, I = ∫(sinx – cosx)/(1 + sinx.cosx) → (1) [Limit: 0 → π/2 ]
Now, applying identity f(x) → f(a + b – x) [where a and b are the limits of the integration]
We have, I = ∫(sin{π/2 – x} – cos{π/2 – x})/(1 + sin{π/2 – x}.cos{π/2 – x})
= ∫(cosx – sinx)/(1 + sinx.cosx) → (2)
Adding (1) and (2)
We have, 2I = 0
or, I = 0
Thanks and regards,
Kushagra
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