integration siikho

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

174 Answers

1
°ღ•๓яυΠ·

61)lnx=t
1/x dx =dt
then u get a std form wich one can solve easily :P

11
Mani Pal Singh ·

Q69

1
°ღ•๓яυΠ·

53)

-x^2=t
2xdx=dt
so ur q is

∫-e^t dt/2

one can int this :P

11
Mani Pal Singh ·

11
Mani Pal Singh ·

Q57

66
kaymant ·

Q-66:


66
kaymant ·


1
rajat sen ·

74.

11
Mani Pal Singh ·


11
Mani Pal Singh ·

QUESTION NUMBER 54

∫02 x31-x2dx

HERE THIS QUESTION COULD ONLY BE DONE BY SUBSTITUTING x=sinθ
OR BY PARTS BUT THERE A TERM OF sin-1x WILL COME

SO THE UPPER LIMIT WILL NOT BE DEFINED
SO ITS INTEGRAL CANNOT BE FOUND

CORRECT ME IF I AM WRONG

11
Mani Pal Singh ·

QUESTION NUMBER 51

put x=4/5tanθ
dx=4/5sec2θdθ

on solving u will get

∫1/4cosecθdθ
=1/4log| cosecθ-cotθ |+C

11
Anirudh Narayanan ·

[3][3]

11
Mani Pal Singh ·

bhai
PLEASE NO BASHING 4 IT
30 POSTS HAVE BEEN DELETED[3][3][3][3]

11
Mani Pal Singh ·

CAN SOMEBODY HELP WITH #131
THAT IS QUESTION NUMBER 54

11
Anirudh Narayanan ·

mani, even if we substitute x2=t,

we get

\frac{1}{2}\int_{0}^{4}{t\; \sqrt{1-t}\; dt}

i don't think we can do this directly so i substituted 1-t = u2

which caused a problem in the limits again (since √1-4 is complex)

13
Двҥїяuρ now in medical c ·

63.∫√tanx-√cotx dx

=∫√2{√sinx +√cosx}/√1-(sinx-cosx)2 dx

=∫dt/√1-t2

[taking sinx-cosx=t ]

13
Двҥїяuρ now in medical c ·

52.∫ dx/x√(5x)2+42

put x=1/t
dx/x2=-dt

I=-∫dt/√52+(4t)2
=-1/4ln(4t+√52+(4t)2)
substitute x

11
Mani Pal Singh ·

ABHIRUP SAID IN #138

52.∫ dx/x√(5x)2+42

put x=1/t
dx/x2=-dt

I=-∫dt/√52+(4t)2
=-1/4ln(4t+√52+(4t)2)
substitute x

BHAI BOLDED SAMAJH MEIN NAHIN AAYA?????

11
Mani Pal Singh ·

65. 1/a∫√1-x2dx-

1st part is by the formula....

2nd part

∫√1-x2dx/x

put 1-x2=z2

∫z2dz/(1-z2)

=-∫dz+∫dz/1-z2
=-z+1/2 ln[(1+z)/(1-z)]

substitute z

MERA BOLDED WAALA f(a+b-x) WALI PROPERTY SE NAHIN BAN RAHA
PLEASE THODA SA EXPLAIN KAR DO

11
Mani Pal Singh ·

@QUESTION NUMBER 54

∫02 x3√1-x2dx

SI ISKO 0 TO 1 AND 1 TO 2 BREAK KARENGE
TO ISS MEIN 1 TO 2 WAALA INTEGRAL DEFINED NAHIN HOGA KYUNKI √1-x2
HAVE TO BE POSITIVE

SO X LIES BETWEEN -1 TO 1

IF THE UPPER LIMIT IS 1 THEN THE QUES CAN BE DONE BY PUTTING
x2=t
2xdx=dt

then

1-t=p
-dt=dp]

so by changing the limits question can be solved

SIR TILL NOW WE HAVE ONLY USD INTEGRATION TO FIND AREA IN REAL PLANE
BUT IF THIS PROBLEMS DEALS WITH ARGAND PLANE

THEN I HAVE NO PROBLEM IN SHOWING THE SOLUTION[1]

13
Двҥїяuρ now in medical c ·

BHAI BOLDED SAMAJH MEIN NAHIN AAYA?????

watz the problem
t=1/x
dt=-dx/x2

11
Mani Pal Singh ·

K
MEINE KYCH AUR SAMJHA THA
BUT PLEASE EXPLAIN #141

13
Двҥїяuρ now in medical c ·

i think i have done a mistake dere...

11
Mani Pal Singh ·

sir please check #141 and clear my doubts !!!!!!!!!!!!!!!

11
Mani Pal Singh ·

ANSWER TO PROBLEM NUMBER 5

1
Lonely 1 ·

\int^{\infty}_0 x / ( e^x - 1) \ dx

question 54

multiply and divide by exand then integrate u will get the answer

39
Dr.House ·

76]\int {\frac{{3x^3 + 5x^2 - 7x + 9}} {{\sqrt {2x^2 + 5x + 7} }}} dx

39
Dr.House ·

77]\int {\frac{{\sqrt[3]{{1 + \sqrt[4]{x}}}}} {{\sqrt x }}} dx

39
Dr.House ·

78]\int {\sqrt {x + \sqrt {x^2 + a^2 } } } dx

39
Dr.House ·

79] \int {\frac{{dx}} {{x\left( {2 + \sqrt[3]{{\frac{{x - 1}} {x}}}} \right)}}}

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