very basic integral

1.evaluate

far practize

11 Answers

1
JOHNCENA IS BACK ·

write in parametric form! denominator becomes sin(x+aplha)

take x+alpha=t and write x=t-alpha.......expand numerator n integrate!

39
Pritish Chakraborty ·

(√cot x) is damn easy job :P

1
" ____________ ·

1. integrate

2.

FOR PRCTC

as alwayz pritish babu has done a nice work !!!!!!!!!!!!!!!!!

1
Manmay kumar Mohanty ·

2) (sinxlogxxcosx)dx=sinxlogxdxxcosxdx .......(1)
let I1 = sinxlogxdx
considering sin x as 2nd and log x as first function we have BY USING BY PARTS,
I1 = logxsinx+xcosx

now substituting the value of I1 in (1) we get,
(sinxlogxxcosx)dx=logxsinx+xcosx - ∫cos xx = logxsinx
= - logx cosx

1
Manmay kumar Mohanty ·

LONG ONE THOUGH.
(tanx+cotx)dx=sinxcosxsinx+cosxdx now put sinx - cosx = t → (cosx + sinx)dx = dt

integral reduces to 21t2dt=2sin1t+c=2sin1(sinxcosx)+c

similarly we find (tanxcotx)dx=2lnsinx+cosx+sin2x+c

now cotx=212cotxdx=21(tanx+cotx)(tanxcotx)dx

subtituting the values of (tanx+cotx)dx and (tanxcotx)dx we get cotx = 21[2sin1(sinxcosx)+2ln(sinx+cosx)+sin2x]+k,k=integ.constant

1
" ____________ ·

excellent work , manmay dada, & rahul

1
sriraghav ·

whats the solution for 1st question ,
iam getting in the for of

K ∫√1- tan α. cos t dt
where t is x - α

1
Unicorn--- Extinct!! ·

@ pritish...u pinked ur post urself?? [5][3]

1
Manmay kumar Mohanty ·

uska paper hi pink hai yaar......

1
arsh sharma ·

cotx
yaar ise poora toh kar...

1
Manmay kumar Mohanty ·

kiya to hai. PRITISH NE AUR MAINE

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