integration of arctan(secx+tanx)
x lying between -pi/2 to pi/2
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The problem is that you divide the numerator and the denominator of the integrand by cos2x, but this can be done only for cos2x≠0 i.e. x≠π/2. So you have to divide your integral in two improper integrals as:
∫π0=∫(π/2)−0−∫(π/2)+π
Your substutition is correct, so you have the limit:
limx→(π/2)±arctan(batanx)
that is ±π/2.
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I hope you are satisfied
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richasharma01:
thankyou i was stuck with (1+sinx)/cosx
is my answer is right?
yes thanks:)
mere questions?
wo integration waale ke answer maine check ki to usmein iota involved hai to wo mujhe nahi aata and and matrix mujhe pasand nahi but try karungi
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