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

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

Grade:12

3 Answers

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
 
Arjun Radhakrishnan
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
therefore the answer is 0
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
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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