Question

What is `\int x/(1-sinx) dx` ?

Answer 1

`I=xtan(pi/4+x/2)-ln|sec(pi/4+x/2)|+c`

Explanation:

Here,

`I=intx/(1-sinx)dx`

`=intx/(1+cos(pi/2+x))dx...to[because cos(pi/2+theta)=-sintheta]`

`=intx/(2cos^2(pi/4+x/2))dx...to[because1+cosalpha=2cos^2(alpha/2)]`

`=int(x/2)/(cos^2(pi/4+x/2))dx`

Let,

`pi/4+x/2=u=>x/2=u-pi/4=>x=2u-pi/2=>dx=2du`

So,

`I=int(u-pi/4)/cos^2uxx2du`

`=int(2u-pi/2)sec^2udu`

`"Using "color(blue)"Integration by Parts :"`

`:.I=(2u-pi/2)intsec^2udu-int[(2-0)intsec^2udu]du`

`I=(2u-pi/2)*tanu-int2tanudu`

`=(2u-pi/2)tanu-2ln|secu|+c`

Subst. back , `u=pi/4+x/2` , we get

`I`=`[2(pi/4+x/2)-pi/2]tan(pi/4+x/2)-2ln|sec(pi/4+x/2)|+c`

`I=[pi/2+x-pi/2]tan(pi/4+x/2)-ln|sec(pi/4+x/2)|+c`

`I=xtan(pi/4+x/2)-ln|sec(pi/4+x/2)|+c`

Answer 2

`I=x(tanx+secx)-ln|secx|-ln|secx+tanx|+C`

Explanation:

Here,

`I=intx/(1-sinx)dx=int(x(1+sinx))/((1-sinx)(1+sinx))dx`

`:.I=int(x(1+sinx))/(1-sin^2x)dx`

`=int(x(1+sinx))/cos^2xdx`

`=intx(1/cos^2x+sinx/cos^2x)dx`

`=intx(sec^2x+secxtanx)dx`

`"Using "color(blue)"Integration by Parts :"`

`color(blue)(intu*vdx=uintvdx-int(u'intvdx)dx`

Let , `u=x and v=(sec^2x+secxtanx)`

`=>u'=1 and intvdx=(tanx+secx)+c`

So,

`I=x*(tanx+secx)-int1(tanx+secx)dx`

`=x(tanx+secx)-{ln|secx|+ln|secx+tanx|}+C`

`=x(tanx+secx)-ln|secx|-ln|secx+tanx|+C`