What is int x/(1+sinx) dx∫x1+sinxdx?

1 Answer
maganbhai P.
Mar 9, 2018

x(tanx-secx)+ln|(secx+tanx)/secx|+cx(tanx−secx)+ln∣∣∣secx+tanxsecx∣∣∣+c

Explanation:

I=intx/(1+sinx)dx=int(x(1-sinx))/(1-sin^2x)dx=int(x(1-sinx))/cos^2xdxI=∫x1+sinxdx=∫x(1−sinx)1−sin2xdx=∫x(1−sinx)cos2xdx
=>I=intxsec^2xdx-intxsecxtanxdx⇒I=∫xsec2xdx−∫xsecxtanxdx
Integration by parts,we get
I=[xtanx-int1*tanxdx]-[x*secx-int1*secxdx]I=[xtanx−∫1⋅tanxdx]−[x⋅secx−∫1⋅secxdx]
I=xtanx-ln|secx|-xsecx+ln|secx+tanx|+cI=xtanx−ln|secx|−xsecx+ln|secx+tanx|+c
=xtanx-xsecx+ln|secx+tanx|-ln|secx|+c=xtanx−xsecx+ln|secx+tanx|−ln|secx|+c
=x(tanx-secx)+ln|(secx+tanx)/secx|+c=x(tanx−secx)+ln∣∣∣secx+tanxsecx∣∣∣+c

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