xtan xProve that:sec x + tan x0

1.	xtan xProve that:sec x + tan x0
Answer» let I = integration of (xTanx / Secx+Tanx) from o to pie -----(1) So using prop of defnite integration (i.e. by putting x = pi -x) we get I = in integration of (pi-x)Tan(pi-x) / Sec(pi-x)+Tan(pi-x) from o to pie which gives,integration of (pi-x)(-Tan(x) / (-Secx -Tanx) from o to pie , OR I = integration of (pi-x)(Tan(x) / (Secx + Tanx) --------(2) Adding (1) &(2) we get 2I = integration ofpi.Tanx / (Secx+Tanx)from o to pie, now convertingtrig termsintosinx &cosx, we get 2I =piX integration of Sinx/(1+ Sinx) from 0 to pi, which gives 2I = pi X integraiton of Secx tanx - tan^2x from 0 to pi, So 2I = piX (Secx -tanx +x) from 0 to pi, OR 2I = pi.((Secpi -tanpi +pi) - (Sec 0 -tan 0 +0)) = pi . ((-1-0+pi) - (1) ) = pi( pi-2), So I = pi .(pi -2) /2

Answer»

let I = integration of (xTanx / Secx+Tanx) from o to pie -----(1)

So using prop of defnite integration (i.e. by putting x = pi -x) we get

I = in integration of (pi-x)Tan(pi-x) / Sec(pi-x)+Tan(pi-x) from o to pie

which gives,integration of (pi-x)(-Tan(x) / (-Secx -Tanx) from o to pie , OR

I = integration of (pi-x)(Tan(x) / (Secx + Tanx) --------(2)

Adding (1) &(2) we get 2I = integration ofpi.Tanx / (Secx+Tanx)from o to pie, now convertingtrig termsintosinx &cosx, we get

2I =piX integration of Sinx/(1+ Sinx) from 0 to pi, which gives 2I = pi X integraiton of Secx tanx - tan^2x from 0 to pi, So

2I = piX (Secx -tanx +x) from 0 to pi, OR

2I = pi.((Secpi -tanpi +pi) - (Sec 0 -tan 0 +0)) = pi . ((-1-0+pi) - (1) ) = pi( pi-2), So

I = pi .(pi -2) /2

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