1Then plz give d hint at least...
HELLO ALL. THIS THREAD IS A REVIVAL OF b555 ' S INTEGRATION SIKHO THREAD.
THAT HAS BEEN A VERY SUCCESSFUL THREAD , PARTICIPATED BY MANY TIITians.
BUT , STILL MANY QUESTIONS OF THAT THREAD ARE REMAINING PENDING.
SO, I'M POSTING THOSE PENDING QUESTIONS HERE , PHASE AFTER PHASE.
SO, HERE STARTS " integration ke liye aur bhi kuch karega "
CONDITIONS APPLIED : kisi ko gali dena ho toh b555 ke chatbox mein dal dena.
so here we go :
1) ∫ x3 / [(x - 1)3 ( x - 2 )]
2)\int_{0}^{\pi /2}{e^x[ cos (sinx). cos^2x/2 + sin(sinx).sin^2x/2]dx}
3)\int_{1}^{e}{[ (1+x)e^x +(1- x)e^-^x]lnx dx}
4)\int_{0}^{\pi }[{cos(rx)dx ] / [ 1- 2acosx+ a^2]
5) \int (tanx)^1^/^3 dx
6)\int_{o}^{infinity}{[lnx / (1-x^2)] dx}
7) In = \int dx / (1+x^2^n)
8) \int sinx / ln(2+x^2)
9) \int_{0}^{\pi /3}{[sin^nx / (sin^nx +cos^nx)]dx}
10) \lim_{n\rightarrow infinity }\sum_{i=1}^{n}{(3/n) sin (2\pi + (3\pi i/n))} let first this much be solved.
-
UP 0 DOWN 0 2 25
25 Answers
116)∫(√cos 2x/sinx)dx
=(√cos 2x.sinx/sin2x)dx
=(√2cos2x-1.sinx/1-cos2x)dx
Put cos x =t so that sin x dx=-dt
=(√2t2-1/t2-1)dt
=[(√2-1/t2)/(t-1/t)]dt
Then what to do...???
Plz help...
1^bro der is only one page
this is the duplicate of integration sekho...
see anoder one started by bhargav
1only one page is showing here ...wht has happened to the thread???????
1girish , here goes ur solution.
I = ∫(√cos 2x/sinx)dx = ∫(cos 2x) / √cos 2x . sinx) dx = ∫ (1 - 2 sin2x) / sinx √cos2x ) dx
= \int dx / sinx \sqrt{cos^2x - sin^2x} - 2 \int sinx dx/\sqrt{2cos^2x-1}
now, solve them independently.
\int dx / sinx \sqrt{cos^2x - sin^2x} = \int cosec^2x dx / \sqrt{cot^2x-1}
now, take cotx = m and solution will be there.
now,
- 2 \int sinx dx/\sqrt{2cos^2x-1} = \int - dn / \sqrt{2n^2-1}
if u take cosx = n i.e. - sinx dx = dn.
thus, now, its integrable.
23its a vry tedious job to type the answers ...and dat too of integration ....LOl
4Sorry I mistakingly deleted the abv post before
I somehow revived it now
411 )INTEG [tan 2θ/(√cos6θ + sin6θ ] Answered
12 ) I = [(5x4+4x5) / (x5+x+1)2 ] dx Answered
13 ) integ [ cos^2xdx/(1+tanx) ] ANSWERED
14 ) ∫ ln[ln(1+x/1-x)] /1-x2 dx ANSWERED
15) ∫ [(x/e)x +(e/x)x] lnx dx ANSWERED
4@aieee I answered 3 of the 5 Qs I posted n 2 of urs already
( In an IIT JEE Paper of 20 qs u can't stop trying 17th qs if 8 th Qs is not answerable !!! )
Anyway , I wont mind to shift my Qs elsewhere if u have a prob in me posting Qs here
415 ) put (x/e)x = t
x ln(x/e) = ln t
x( ln x - 1) = ln t
differentiate
(ln x +1 -1 ) dx = dt/t
ln x dx = dt/t
this reduces our integrand to (t +1/t)/t dt
i.e (1 + 1/t2 )dt
on integrating, it gives t - 1/t + c
on putting back the value of t i.e. (x/e)x
our answer is (x/e)x - (x/e)-x +c Answer
1UTTARA , try answering the questions !!!!!!!!!!!!!
don't just keep on adding questions . dat cn be done by anyone !
If response is not coming , then i believe tiitians need some more time to catch up wid this thread.
413 )cos^2xdx/(1+tanx)
multiply and divide by sec^4x
sec^2xdx/(sec^4x)(1+tanx)
tanx=t
sec^2xdx=dt
dt/(1+t)(1+t^2)^2
now solve by using partial fraction,,,
3q4)\int_{0}^{\pi }{\frac{cosx}{1-2acosx+a^{2}}}dx=\int_{0}^{\pi }{-cosx/1+2acosx+a^{2}}dx
rest is adding and integrating.
1Q.7)
http://targetiit.com/iit-jee-forum/posts/for-all-u-iitians-out-there-even-my-fiitjee-sir-co-10832.html
http://targetiit.com/iit-jee-forum/posts/indefinite-integration-pls-help-10786.html
13)\int_{1}^{e}{[ (1+x)e^x +(1- x)e^-^x]lnx dx}
this is perhaps not in IIT syllabus. which i think is a type of double integration.
separate the terms i.e. ∫(1+x)ex lnx dx + ∫(1-x)e-x lnx dx
solving independently for ∫(1+x)ex lnx dx , need to take :
u = ex and v=( x+1) lnx , then du/dt and dv/dt and putting both in the expression.
1seems this thread is not attracting crowd. k. let me post some solutions , in a hope of keeping this thread alive.
5) I = ∫ (tanx)1/3 dx.
Let tanx = m3/2 , then sec2x dx = 3/2 m1/2
now I = ∫ \sqrt[3]{m^3^/^2 }. (3/2) m^1^/^2 . ( 1/sec^2x)dm
=3/2 \int (m dm)/ (1+m)(1-m+m^2)
now,let : m/(1+m)(1-m+m^2) = A/(1+m) + (Bm+C) / (m^2-m+1)
solving it , we get A=-1/3 , B=1/3 , C=1/3.
SO , I = (3/2)[\int -1/3. dm/(1+m) + \int (m/3+1/3)/ (m^2-m+1)
now, its integrable form.
23Q11] ( i m taking theta as A]
I = ∫ tan 2A dA
√sin6A + cos6A ..(1)
sin6A + cos6A
= (sin2A)3 + (cos2A)3
= (sin2A+cos2A)(sin4A - sin2Acos2A + cos4A)
= ( sin4A + cos4A - sin2Acos2A)
= ( [ (sin2A+cos2A)2 - 2 sin2Acos2A] - sin2Acos2A )
= ( 1 - 3 sin2Acos2A )
= ( 1 - 4sin2Acos2A + sin2Acos2A )
= ( 1 - sin22A + sin22A )
4
=( cos22A + sin22A )
4
=cos22A ( 1 + tan22A/4 )
=cos22A ( 3/4 + sec22A/4 ) .....(2)
frm (1) and (2)
I = ∫ tan 2A dA
cos2A √( 3/4 + sec22A/4 )
= ∫ tan 2A sec2A dA
√( 3/4 + sec22A/4
= ∫ 2tan 2A sec2A dA
√( 3 + sec22A
put sec2A = t
2sec2Atan2A = dt
and solve ......[ ignore silly mistakes plz ]
4Ans 14 ) Put ln (1 +x)/ (1-x) = t
dt/dx = 2/(1-x2)
I= - 1/2 integ [ lnt ]
= -1/2 t (lnt + 1 ) + c
Now substitute for t
11Ans12) I = [(5x4+4x5) / (x5+x+1)2 ] dx
I =[ x4(5+4x) / x10 (1+ 1/x4 + 1/x5)2 ] dx
I = (5/x6 + 4/x5) / (1+ 1/x4 + 1/x5)2 dx
Now put 1+1/x4 + 1/x5 = t
therefore, (-4/x5 - 5/x6 )dx = dt
I = ∫-dt / t2 = 1/t+c = 1 / (1 + 1/x4 + 1/x6) + c
Therefore, I = x5 / (x5+x+1) + c
39kindl use the necxt numbers while posting new questions. i mean uttara u could have continued with 11,12,13,........
anyways edit that post. that would make the users viewing this thread in fuyure nt to get confused
41) Split into partial fractions
x3/(x-1)3(x-2) = A/(x-2) + B/(x-1) + C/(x-1)2 + D/(x-1)3
46) Integ (0 -> ∞) ln x / x(1/x - x) dx
Put ln x = t => 1/x dx = dt
I = integ dt/(e-t - et)
= 1/2 [ ln (-e2t ) - ln (1-e2t) ]
Substitute for t
3q5)http://targetiit.com/iit-jee-forum/posts/an-integral-difficult-to-resist-10816.html