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[(tanx)/x].dx

[(tanx)/x].dx

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3 Answers

Parth Shrivastava
34 Points
9 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

jbdjhsfgbr gnjsndvjesr
16 Points
9 years ago

its a very easy question... try teaching something more useful to students....Frown

Anantha padmanabam
26 Points
6 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 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)

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

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