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# [(tanx)/x].dx

8 years ago

This can be integrated by integration by parts
let 1/x=d v
then logx dx =v on integrating
let tanx=u
sec^2xdx=du
udv = uv-  vdu = logxtanx-  logxsec^2xdx
so tanx/x dx= logxtanx-  ∫ logx sec^2xdx------ (1)

now consider the second term in the RHS
logx sec^2x dx, again applying by parts technique
let u= logx , du= 1/xdx sec^2x dx= dv , v= tanx on integrating

so this becomes logxtanx- tanx/x dx

substituting in eqn (1)
integral tanx/x dx= logxtanx+logxtan-   tanx/xdx
2 tanx/x dx= 2logxtanx

so tanx/x =logx tanx+c

Ans

8 years ago

its a very easy question... try teaching something more useful to students.... 4 years ago

let 1/x=d v
then logx dx =v on integrating
let tanx=u
sec^2xdx=du
udv = uv-  vdu = logxtanx-  logxsec^2xdx
so tanx/x dxlogxtanx-  ∫ logx sec^2xdx------ (1)

now consider the second term in the RHS
logx sec^2x dx, again applying by parts technique
let u= logx , du= 1/xdx sec^2x dx= dv , v= tanx on integrating

so this becomes logxtanx- tanx/x dx

substituting in eqn (1)

After substituting in eqn 1 u will get logx tanx - logx tanx + internal (tanx/x)
Which will give back the question