# Integral of (sinx-cosx)/(1+sinxcosx) with limits zero to pi/2

Chandan
121 Points
6 years ago
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).
=$\tan^{-1}$((t-1/2)/1/2)                                               ∫da/(a2+k2)=$\tan^{-1}$(a/k)
=$\tan^{-1}$(2t-1)+c
=$\tan^{-1}$((2tanx-1))+c

11 Points
6 years ago
$\int_{0}^{\pi /2}(sin x-cos x)/(1+ sin xcos x)$        = I
change x=($\pi /2$​  –  x)

then you will get
$\int_{0}^{\pi /2}(cos x-sinx)/(1+ sinx cosx)$   =   I
add both of them(i.e I+I =2I)
2*I = $\int_{0}^{\pi /2}(0/(1+ sinxcosx))$

2*I=0
I = 0
4 years ago
Dear student,

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)