11ashish recheck ur solution
the answer coming out is 3
by limit as a sum
ur 2nd step is doubtful to me
this thread is only for integration sums............
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1}\int_{0}^{1}e^{\sqrt{e^{x}}}\ dx+2\int_{e}^{e^{\sqrt{e}}}\ln (\ln x)\ dx
2} \int \frac{x^{3}}{(x-1)^{3}(x-2)}\ dx
3}1985 japan women university
\lim_{a\rightarrow + \infty} \frac {\int_0^a \sin ^ 4 x\ dx}{a}
4} \frac{1}{\displaystyle \int _0^{\frac{\pi}{2}} \cos ^{2006}x \cdot \sin 2008 x\ dx}
5} \int_0^{\frac {\pi}{2}} \frac {x^2}{(\cos x + x\sin x)^2}\ dx
6} IF F(X)= x2 + |x| then prove that
\int_{0}^{\pi}f(\cos x)\ dx=2\int_{0}^{\frac{\pi}{2}}f(\sin x)\ dx
7}Evaluate the following definite integral.
\int_{e^{2}}^{e^{3}}\frac {\ln x\cdot\ln (x\ln x)\cdot\ln\{x\ln (x\ln x)\} + \ln x + 1}{\ln x\cdot\ln (x\ln x)}\ dx
8}
Let f a nonnegative ,continuous and periodical function defined on the reals, such that the arithmetic mean of the numbers f(1), f(2),...f(n) tends to zero when n tends to infinity. Prove that f(k)=0 for any natural number k.
9}\int_{0}^{1}\frac {x\ dx}{(x^{2} + x + 1)^{\frac {3}{2}}}
10}\int_{2}^{6}\ln\frac { - 1 + \sqrt {1 + 4x}}{2}\ dx
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174 Answers
1111HAVE A LOOK AT IT
http://targetiit.com/iit_jee_forum/posts/important_information_2774.html
1hey dude isko bhi integrate karo
∫ (√x-3 ) ( sin-1(ln(x) ) + cos-1(ln(x) ))
118)1.> take x=t2
the limits remain as 0 and 1
dx=2t.dt and the rest is the integral of sec-1(x)
62f(x) = \int_{2}^{e^x}\frac{1}{\sqrt{\ln t}}dt
k=lnt\\ \\ t=e^{k}\\ dk=frac{1}{e^{k}}dt\\ \\ dt={e^{k}}dk\\
does this help?
1hey guys do tht √1+x3 wala naa........its difficult and different........wel cubic under a root is dere for our syllabus????
11HAVE A LOOK AT IT
http://targetiit.com/iit_jee_forum/posts/important_information_2774.html
66The integration of √1+x3 cannot be written in terms of the elementary functions. It will come in terms of the Elliptical integral.
66Q4) Ans: 2007
Hint: Write the denominator as \int_0^{\pi/2} \cos^{n-1}x \,\sin (n+1)x \ \mathrm{d}x. Exapand the sine to get:
\int_0^{\pi/2}\cos^{n-1}x \big(\sin nx \cos x + \cos nx \sin x\big)\ \mathrm{d}x\\[2ex] =\int_0^{\pi/2}\cos^nx\,\sin nx \ \mathrm{d}x + \int_0^{\pi/2}\cos nx\ \cos^{n-1}x \sin x \ \mathrm{d}x
Now integrate the second one by parts.
1Bhargav.. sorry for the behavior that day..
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11THANX BUDDY 4 UR KIND HELP [1]
IT WAS VERY ESSENTIAL 4 THIS THREAD
21q20]
LET lnt = x2
1/tdt = 2xdx
I = ∫ex*2x/xdx
I = ∫exdx over appropriate limitsssss
66First determine the indefinite integral:
I=\int\dfrac{x\cos x}{(x\sin x+\cos x)^2}\ x\sec x\ \mathrm{d}x
Note that \int\dfrac{x\cos x}{(x\sin x+\cos x)^2}\ \mathrm{d}x = \int \dfrac{\mathrm{d}(x\sin x + \cos x)}{(x\sin x+\cos x)^2}=-\dfrac{1}{x\sin x+\cos x}
Therefore,
I = x\sec x \left(-\dfrac{1}{x\sin x+\cos x}\right) + \int \dfrac{\sec x(1+x\tan x)}{x\sin x+\cos x}\ \mathrm{d}x
= \dfrac{-x\sec x}{x\sin x+\cos x} + \int \sec^2x\ \mathrm{d}x
= \dfrac{-x\sec x}{x\sin x+\cos x} + \tan x=\dfrac{\sin x-x\cos x}{x\sin x+\cos x}+C
Hence the definite integral is
\left|\dfrac{\sin x-x\cos x}{x\sin x+\cos x}\right|_0^{\pi/2}=\dfrac{2}{\pi}
66We have
\mathrm{d}\big(x\ln\{x\ln(x\ln x)\}-x\big)
=\ln\{x\ln(x\ln x)\}\ \mathrm{d}x+ x \cdot \dfrac{1}{x\ln(x\ln x)}\left(\ln(x\ln x)+ x\cdot \dfrac{1}{x\ln x}\left\{\ln x+ 1\right\}\right)\ \mathrm{d}x-\mathrm{d}x
=\left(\ln\{x\ln(x\ln x)\}+\dfrac{\ln(x\ln x)+\dfrac{1+\ln x}{\ln x}}{\ln(x\ln x)}-1\right)\ \mathrm{d}x
=\dfrac{\ln x \cdot \ln(x\ln x)\cdot \ln\{x\ln(x\ln x)\}+\ln x+1}{\ln x\cdot \ln(x\ln x)}\ \mathrm{d} x
Therefore,
\int\dfrac{\ln x \cdot \ln(x\ln x)\cdot \ln\{x\ln(x\ln x)\}+\ln x+1}{\ln x\cdot \ln(x\ln x)}\ \mathrm{d} x = x\ln\{x\ln(x\ln x)\}-x +C
The definite integral can now be easily evaluated.
1145.
lim n → ∞ (1/n!)^(1/n)
Since it is raised to 1/n. Here 1/∞ = 0
Therefore the ans is 1
1∫ (√1+x4)dx
(1-x4 )
And haan answer in x and functions only!
Ya this sum is xeroxed one so dont be surprised if u si dis in a book !
:-D
1hey dude isko bhi integrate karo
∫ (√x-3 ) ( sin-1(ln(x) ) + cos-1(ln(x) ))
@ manipal singh
dude ye to trap questoin tha ,just check ki the function exist or not[1]
