Integral Calculus> indefinite integrals...
3 AnswersLast Activity: 14 Years ago
1) I=tan^-1secx+tanx)dx
secx+tanx=1+sinx/cosx
1+sinx=(sinx/2+cosx/2)^2
cosx=(cosx/2-sinx/2)(cosx/2+sinx/2)
I=tan^-1(sinx/2+cosx/2)/(cosx/2-sinx/2)dx
=tan^-(tanpi/4+x/2)dx
=(pi/4+x/2)dx =(pi/5)x + x^2/4 +k
2)I=(1+2tanx(tanx+secx)^1/2dx
=(1+tan^2x + tan^2x +2secxtanx)^1/2 dx
=( sec^2x +tan^2x +2tanxsecx)^1/2dx
= ((secx+tanx)^2 )^1/2dx
= ( secx+tanx )dx
=logsecx+tanx +logsecx +c
=log(secx+tanx)secx +c
Last Activity: 14 Years ago
=tan-1(1+sinx/cosx)
=tan-1((sinx+cosx)2/(cosx2-sinx2))
=tan-1(tan(x+∏/4)
=x+pi/4
Last Activity: 4 Years ago
I=integ.of. tan^-1(1+sinx)/cosx.dx
I=integ.of. tan^-1(cos^2x/2+sin^2x/2+2.sin x/2.cos x/2).dx /(cos^2x/2-sin^2x/2)
I=integ.of tan^-1(cos x/2+sin x/2)^2.dx/(cos x/2+sin x/2).(cos x/2-sin x/2).
I=integ.of tan^-1(cos x/2+sin x/2).dx/(cos x/2-sin x/2).
I=integ.of tan^-1 (1+tan x/2)/(1-tan x/2).dx
I=integ.of tan^-1 (tanπ/4+tan x/2)/(1-tan π/4.tan x/2).dx
I=integ.of tan^-1 tan (π/4+x/2).dx
I=integ.of. (π/4+x/2).dx
I = π/4. x. +x^2/(2.2). +C
I= π.x/4. +x^2/4. +C. Answer.
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