Create your Smart Home App in the Free Webinar on Home Automation. Register Now
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Free webinar on App Development Learn to create your own Smart Home App
16th Jan @ 5:00PM for Grade 1 to 10
Sit and relax as our customer representative will contact you within 1 business day
[(tanx)/x].dx
This can be integrated by integration by partslet 1/x=d vthen logx dx =v on integratinglet tanx=usec^2xdx=du∫ udv = uv- ∫ vdu = logxtanx- ∫logxsec^2xdxso ∫tanx/x dx= logxtanx- ∫ logx sec^2xdx------ (1) now consider the second term in the RHS∫ logx sec^2x dx, again applying by parts techniquelet 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/xdx2 ∫tanx/x dx= 2logxtanx so ∫tanx/x =logx tanx+c Ans
This can be integrated by integration by partslet 1/x=d vthen logx dx =v on integratinglet tanx=usec^2xdx=du∫ udv = uv- ∫ vdu = logxtanx- ∫logxsec^2xdxso ∫tanx/x dx= logxtanx- ∫ logx sec^2xdx------ (1)
now consider the second term in the RHS∫ logx sec^2x dx, again applying by parts techniquelet 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/xdx2 ∫tanx/x dx= 2logxtanx
so ∫tanx/x =logx tanx+c
Ans
its a very easy question... try teaching something more useful to students....
let 1/x=d vthen logx dx =v on integratinglet tanx=usec^2xdx=du∫ udv = uv- ∫ vdu = logxtanx- ∫logxsec^2xdxso ∫tanx/x dx= logxtanx- ∫ logx sec^2xdx------ (1) now consider the second term in the RHS∫ logx sec^2x dx, again applying by parts techniquelet 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
let 1/x=d vthen logx dx =v on integratinglet tanx=usec^2xdx=du∫ udv = uv- ∫ vdu = logxtanx- ∫logxsec^2xdxso ∫tanx/x dx= logxtanx- ∫ logx sec^2xdx------ (1)
substituting in eqn (1)
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -