[(tanx)/x].dx

Question

[(tanx)/x].dx

Parth Shrivastava · Answer

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 · Answer

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

Anantha padmanabam · Answer

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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[(tanx)/x].dx

[(tanx)/x].dx

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