39so u have been going to mathlinks these daya....good
have u registered??
Q 1) \int\frac{x-1}{(x+1)\sqrt{x^{3}+x^{2}+x}}dx
Q 2) \int{\left({\tan^{-1}\sqrt{(\sqrt x-1)}}\right)}dx
Q 3) \int{(x^{1/3}+(\tan x)^{1/3})}dx
Q 4) \int_{1}^{e}\frac{Inx}{x(\sqrt{1-Inx}+\sqrt{Inx+1})}dx
Q 5) \int_{\frac{\pi }{2}}^{\frac{\pi }{4}}\frac{1}{1+sin 2x}\sqrt[2009]{\frac{tanx-1}{tanx+1}}dx
These are all from AOPS...
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10 Answers
39
62no not mathlinks but AOPS.. are tehy the same? I have no time otherwise would want to register and use them more...
39yes both are the same ,
mathlinks is of damain .ro and aops is .com. its a matter of just seeing the cntents as u want..
u have a bit of advertising in aops , which u wont find in www.mathlinks.ro , otherwise they both are pretty uch same..
all other content remains same
62oh i din know that... :)
but it is a great site for advanced users ... I wish a day had 50 hours :P
11Ans2) Putting x = sec4 @ and dx = 4 sec 4 @ tan @ d@
I= ∫ tan -1 (tan @) 4 sec 4 @ tan @ d@
= 4 ∫ @ sec 4 @ tan @ d@
= 4 ∫ @ sec 3 @ sec @ tan @ d@
= 4 ∫ @ sec 3 @ d(sec @)
Now by using partial fraction, we get
= 4 { @ sec 4 @ / 4 - (1/4) ∫ sec 4 @ d@ }
= @ sec 4 @ - ∫ sec 4 @ d@
= @ sec 4 @ - ∫ (1 + tan 2 @ ) sec 2 @ d@
= @ sec 4 @ - tan @ - tan 3 @ / 3 +c
= @ sec 4 @ - √(sec 2 @ - 1) - (1/3) (sec 2 @ -1 ) 3/2
=x tan -1 (√(√x -1 ) ) - (√(√x -1 ) ) - (1/3) (√x - 1 ) 3/2 + c
11Ans1) Multiplying the numerator and denominator by (x+1)
I = ∫ (x2 -1 ) dx / ( x+1) 2 √(x3+x2+x)
I = ∫ (x2 -1 )dx / (x2+2x+1) √(x3+x2+x)
On deviding numerator and denominator by x2
I = ∫ (1 - 1/x 2 ) dx / (1 + 1/x + 2) (x + 1/x + 1 ) 1/2
Putting (x + 1/x +1 ) = t 2 therefore, 1 - 1/x2 dx = 2t dt
I = ∫ 2t dt / (t2+1) t
I = 2 ∫ 1 dt / (t 2+1) = 2 tan -1 (t) + c
Therefore, I = 2 tan -1 (x+ 1/x + 1) 1/2 + c
106Q4. \int_{1}^{e}{\frac{lnx}{x(\sqrt{1-lnx}+\sqrt{lnx+1})}}
take lnx = t
x-->1, t--> 0
x--> e, t--> 1
\int_{0}^{1}{\frac{t}{(\sqrt{1-t}+\sqrt{t+1})}}
= \int_{0}^{1}{\frac{t(\sqrt{t+1}-\sqrt{1-t})}{2t}}
this can now easily be evaluated
13)∫(x1/3+(tanx)1/3)dx
=3x4/3/4 +∫(tanx)1/3dx
take tanx =t3 & proceed
we get 3∫(t2t/(1+(t2)3)dt
now put t2=z
we get 3/2∫z/(1+z3)dz.......solve further by partial fraction
39sindhu its not that easy...its not like sqrt tanx. I tried it.
1708Ans:(5) \int_{\pi/2 }^{\pi /4}{(tanx-1/tanx+1)^{1/2009} }*(1/(1+sin2x)) here (1/(1+sin2x))= 1/(sinx+cosx)^{2}= sec^{2}x/(tanx+1)^{2} and then put (tanx-1/tanx+1)=t^{2009} and rest of the work is easy.