1^bro der is only one page
this is the duplicate of integration sekho...
see anoder one started by bhargav
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.
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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
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