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Integration limit [0-Pie] (xtanx)/(secx+cosx)

Sunny , 11 Years ago
Grade
anser 1 Answers
Rinkoo Gupta
let I=integralo [0 to pi] x tanx/(secx+tanx)dx
=integral [0 to pi] (pi-x)tan(pi-x)/sec(pi-x)+tan(pi-x) dx
=pi integral[0 to pi] tanx/secx+tanx dx -I
=>2I=pi integral [0 to pi] tanx/(secx+tanx) dx
=>I=pi/2 Integarl [0 to pi] tanx/(secx+tanx) dx
=pi/2 integral (sinx/cosx)/(1+sinx)/cosx
=pi/2 integral [0to pi] sinx/1+cosx
=pi/2 integral (sinx+1 -1)/1+sinx dx
=pi/2[ integral (0 to pi) dx-integral (0 to pi) dx/(1+sinx)]
=pi/2[ pi- integral (0 to pi) (1-sinx)/(1+sinx)(1-sinx)]
=pi/2 [pi-integral (0 to pi) (1-sinx)/cos^2x dx]
=pi/2 [pi-integral (0 to pi) sec^2xdx + integral ( 0 to pi) secxtanx dx]
=pi/2 [pi- (tanx) 0 to pi + (secx) 0 to pi]
=pi/2 [ pi-tan(pi)+tan(0) +sec(pi)-sec(0)]
=pi/2 [pi-1-1]
=pi/2(pi-2) Ans.
Thanks & Regards
Rinkoo Gupta
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Last Activity: 11 Years ago
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