# 1.integration of tan^-1(secx+tanx)2.tan^-1(2cos^2theta/2-2-sin2theta)sec^2theta2.integration of (1+2tanx(tanx+secx)^1/2

Grade:6

## 3 Answers

vikas askiitian expert
509 Points
12 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

aman bansal
askIITians Faculty 33 Points
12 years ago

=tan-1(1+sinx/cosx)

=tan-1((sinx+cosx)2/(cosx2-sinx2))

=tan-1(tan(x+∏/4)

=x+pi/4

ankit singh
askIITians Faculty 614 Points
2 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.