integration siikho

this thread is only for integration sums............

AND YA FORGOT TO TELL ALL TARGETIT USERS, I HAVE COPIED ALL THESE SUMS. SO KISIKO GAALI DENA HAIN YA KUCH KEHNA HAIN, PLEASE VISIT

http://targetiit.com/profile711.html AND POST THEM IN CHATBOX. THANKS IN ADVANCE............

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

39
Dr.House ·

11} \int_0^{\frac {\pi}{2}} e^x\left\{\cos \ (\sin x)\cos ^2 \frac {x}{2} + \sin \ (\sin x)\sin ^ 2 \frac {x}{2}\right\}\ dx

12} \int_{ - \pi}^{\pi} \frac {\sin nx}{(1 + 2009^x)\sin x}\ dx\ (n = 0,\ 1,\ 2,\ \cdots)

13] \int_0^1 \sqrt {\frac {x + \sqrt {x^2 + 1}}{x^2 + 1}}\ dx

14} \int_1^e \{(1 + x)e^x + (1 - x)e^{ - x}\}\ln x\ dx

15} \dislaystyle \left|\frac {\int_0^{\frac {\pi}{2}} (x\cos x + 1)e^{\sin x}\ dx}{\int_0^{\frac {\pi}{2}} (x\sin x - 1)e^{\cos x}\ dx}\right|

16} \int_{0}^{\pi}\frac{\cos (rx)}{1-2a\cos x+a^2}\;dx

17]\int_{1}^{e} \frac{(1 + \ln{x})^2}{x} dx

18}\mathop \int_0^{1}\frac{1}{x(1-x)^\frac{1}{2}}dx

19} \mathop \int_0^{1}\frac{x-1}{lnx}dx

20} f(x) = \int_{2}^{e^x}\frac{1}{\sqrt{\ln t}}dt

39
Dr.House ·

21]\int \frac{x+\sin ^ 2 x}{x\sin ^ 2 x}dx

22]\int \frac{x+\cos 2x +1}{x\cos ^ 2 x}dx

23]\int a^{-\frac{x}{2}}dx\ \ (a>0,a\neq 1)

24]\int \frac{\sin ^ 3 x}{1+\cos x}dx

25]\int_0^{+\infty}\bigg\{\frac 1{4\sinh^2\frac x2}-\frac 1{x^2}\bigg\}\ln x\,dx

26]\int {\sin ^{ - 1}( x)} dx \\

27]\int {y^3 \sqrt{y^2 - a^2 } dy} \\

28]\int_{0}^{\infty} \frac{ln(x)}{1-x^2}dx

29]f(x) = x^{4}+ \left|x \right| I_1=\int_0^\pi f(\cos x)\ dx,\ I_2=\int_0^\frac{\pi}{2} f(\sin x)\ dx. find \frac{I_{1}}{I_{2}}

30]I_n = \int {\frac{1}{{1 + x^{2n} }}dx}

39
Dr.House ·

31]\int \frac{1}{x-1+\sqrt{x+1}} dx

32}[\int \frac{sinx}{\ln (2+x^{2}) }

33]\int_0^{\pi/3} \frac{sin^{n}x}{sin^{n}x+cos^{n}x} dx

34]\int {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}dx}

35]\int {\frac{{dx}}{{x(1 + 2\sqrt x + \sqrt[3]{x})}}}

36]For t>0, find the minimum value of \int_{0}^{1}x|e^{-x^{2}}-t| dx.

37]Prove the following inequality for x≥0

\int_{0}^{x}(t-t^{2})\sin^{2004}t\ dt <\frac{1}{2006}

38]Let f(x) be the function defined for x≥0 which satisfies the following conditions.

a)f(x)=\begin{cases}x \ \ \ \ \ \ \ \ ( 0\leq x<1) \\ 2-x \ \ \ (1\leq x <2) \end{cases}

b) f(x+2n)= f(x) (n=0,1,2,3,...................)

find \lim_{n\to\infty}\int_{0}^{2n}f(x)e^{-x}\ dx.

39]\displaystyle\lim_{n\to\infty}\displaystyle\sum_{i = 1}^{n}\frac {1}{n}\cos \left(\frac {\pi i}{2n}\right)39]

40]\displaystyle\lim_{n\to\infty}\displaystyle\sum_{i = 1}^{n}\frac {3}{n}\sin \left(2\pi + \frac {3\pi i}{n}\right)

39
Dr.House ·

41]\displaystyle\lim_{n\to\infty}\displaystyle\sum_{i = 1}^{n}\frac {1}{n + i}

42]\displaystyle\lim_{n\to\infty}\displaystyle\sum_{i = 1}^{n}\frac {2}{n}\left(2 + \frac {2i}{n}\right)^5

43]\displaystyle\lim_{n\to\infty}\displaystyle\sum_{i = 1}^{n}\frac {2}{n}\left(1 + \frac {2i}{n}\right)

44]\displaystyle\lim_{n\to\infty}\displaystyle\sum_{i = 1}^{n}\left(\frac {4}{n} + \frac {32i}{n^2} + \frac {64i^2}{n^3}\right)

45]lim_{n - > \infty}{(1/n!)^{1/n}}

46]\int _0^1 \frac {x^b - 1}{\ln x}dx = ?

47] suppose that f(x) = 0 for all but finitely many points x in [a,b]. Show that f is integrable on [a,b].

48}

http://www.haverford.edu/math/rmanning/teaching_materials/candysched114.pdf

see to that

49]Two discs, each of radius=a are placed one on top of the other. The disc on the top begins to move with a velocity=v such that v α 1/A`, where A` is the area exposed of the stationary disc underneath. Find time taken by the moving disc to completely expose the disc underneath

that bolded part means v is inversely proportional to 1/A`

50] prove that \int sin^{3/2}2\theta sin \theta d \theta = 2 \sqrt{2} \int sin^{5/2} \theta cos ^{3/2} \theta d \theta

13
Двҥїяuρ now in medical c ·

5]∫x2dx/(cosx+xsinx)2

=∫xcosx xsecx dx/(cosx+xsinx)2
integrating by parts

and on simplification

=xsecx∫d(xsinx+cosx)/(cosx+xsinx)2-∫sec2x(xsinx+cosx)dx/xsinx+cosx

=-xsecx/xsinx+cosx + tanx + C

106
Asish Mahapatra ·

6) f(cosx) = cos2x + lcosxl
= cos2x + cosx .... x ε [0,π/2]
= cos2x - cosx .... x ε [π/2,π]
so. ∫f(cosx) from 0 to pie/2 is ∫(1+cos2x)/2dx + ∫cosxdx
= [x + sin2x/2]/2 + sinx
= π/4 + 1

∫f(cosx) from pie/2 to pie is ∫(1+cos2x)/2dx - ∫cosxdx
= [x + sin2x/2]/2 - sinx
= π/2 - (π/4-1)
= π/4 + 1

so ∫f(cosx)dx from 0 to pie = π/2 + 2 ...... (i)

f(sinx) = sin2x + lsinxl
= sin2x + sinx .... x ε [0,π/2]
so, ∫f(sinx)dx from 0 to pie/2 is ∫(1-cos2x)/2dx + ∫sinxdx
= [x - sin2x/2]/2 - cosx
= [Ï€/4] - [-1]
= π/4 + 1 ..... (ii)

from (i) and (ii) (i) = 2.(ii)
so, ∫f(cosx) = 2.∫f(sinx) limits are given in question...

106
Asish Mahapatra ·

(3) ∫sin4dx = ∫(1-cos2x)2dx/4 = ∫(1-2cos2x+cos22x)dx/4
= ∫[1-2cos2x+ (1+cos4x)/2]dx/4
= ∫(2-4cos2x+1+cos4x)dx/8
= ∫(1-4cos2x+cos4x)dx/8
= [x - 2sin2x + (sin4x)/4]/8
taking limits as 0 and a
= [a-2sin2a+(sin4a)/4]/8
= (4a - 8sin2a + sin4a)/32

now, lim(a→∞)[∫sin4dx]/a = lim(a→∞)[4-8sin2a/a + sin4a/a]/32
= 1/8............

1
Akand ·

17..... take 1+lnx as t
so 1/xdx=dt
there fore it will be 1∫2t2dt
=t3/3+c
=8/3-1/3 = 7/3........

1
Akand ·

24.I=∫(1-cos2x)sinxdx/1+cosx
=∫(1-cosx)sinx/dx
take 1-cosx =t
sinxdx=dt
so I=∫tdt=t2/2+c
=(1-cosx)2+c

11
Subash ·

24

I=\int (8sin^3x/2cos^3x/2)/2cos^2x/2

then take sinx/2 =t

1
Akand ·

26... I=∫1.sin-1xdx
ILATE........
I= xsin-1x-∫x/√1-x2dx
let 1-x2=t2
so -2xdx=2tdt
I=xsin-1x+∫tdt/t
I=xsin-1x+√1-x2+c

1
Akand ·

29.
f(cosx)=cos4x+cosx ........for 0 to ∩/2
f(cosx)=cos4x-cosx..........for∩/2 to ∩
0∫∩=0∫∩/2+∩/2∫∩
solving this.........
..........
I1=3∩/4+2
and I2=∩/4-3/8
wel i dunno.......sum stupid answwer
I1/I2=6∩+16/2∩-3........
IM DAMN SURE THIS ISNT D ANSWER

1
Akand ·

30.......for n=1
I=tan-1x+c....................hehe

1
Akand ·

23....
y=∫a-x/2dx
logy=-loga/2∫xdx........I dunno whether i can do this..
logy=-x2loga/4+c
logy=loga-x2/4+c
y=eloga-x2/4+c
y=a-x2/4.ec

1
Akand ·

27.
I=y2√y2-a2ydy
take y2-a2=t2
ydy=tdt
I=(t2+a2)t.tdt
I=(t4+a2t2)dt
I=t5/5+a2t3/3+c
I=( y2-a2)2√y2-a2/5+ (y2-a2)√y2-a2/3+c

1
Akand ·

29.
SORRY MADE A MISTAKE.

f(cosx)=cos4x+cosx ........for 0 to ∩/2
f(cosx)=cos4x-cosx..........for∩/2 to ∩
0∫∩=0∫∩/2+∩/2∫∩
solving this.........
..........
I1=3∩/8
and I2=3∩/16
I1/I2=2

11
virang1 Jhaveri ·

21
∫(x+sin2x)dx/xsin2x
∫dx/x + ∫dx/sin2x
log x + ∫dx/sin2x
log x + ∫cosec2xdx
log x - cot x

Thx Bhaiyya

11
virang1 Jhaveri ·

22).
∫( x + cos 2x + 1)dx/xcos2x
∫(x +2cos2x -1 +1)dx/xcos2x
∫(x + 2cos2x)dx/xcos2x
∫dx/cos2x +∫2dx/x
tan x +2 log x + c

1
Akand ·

ok 35....
let x=t6.....
so dx=6t5dt...
I=6t5dt/t6(1+2t3+t2)
I=6dt/t(1+2t3+t2)
partial frac.......A=6,B=-6
I=6dt/t-6dt/(1+2t3+t2)
I=6logt-6dt/(1+2t3+t2)+c
and im stuck after this........

1
Akand ·

I=6logt-3/2log(1+t)-∫9dx/2(2t2-t+1)
I=6logt-3/2log(1+t)-9/2(∫(t+1)dt/2t2-t+1-∫tdt/2t2-t+1)..............man i cant understand

1
Akand ·

but its imaginary..............................so can I???? then its simple......sorry

62
Lokesh Verma ·

2t2-t+1

t2-t/2+1/2

(t-1/4)2+ .....

now is this done?

1
Akand ·

ya ya bhaiyya...........its simple........i got it................but shud i type d answer???

1
Akand ·

I=6logt-3/2log(1+t)-9/2(∫(t+1)dt/2t2-t+1-∫dt/2t2-t+1)
I=I=6logt-3/2log(1+t)-9/2(∫1/4(4t-1)dt/(2t2-t+1)+5/4∫dx/(2t2-t+1))-∫dt/2t2-t+1)..........
and rest solve as nishu bhaiyya said........I cant type anymore....lazy and confused and bored enuf........

11
Mani Pal Singh ·

21
tapanmast Vora ·

Q24)

deno = 2 cos2(x/2);

num = 8 sin3(x/2) cos3(x/2)

ther4 fractn = 4sin3(x/2)cos(x/2);

now let sin(x/2) = t;
.5*cos(x/2)dx = dt

ther4 ∫8t3 dt;

2 sin(x/2)4 + c

21
tapanmast Vora ·

34]
RATIONALISE N THEN SPLIT;

ANS : 4√x - x - ln(1+√x) + c

11
Mani Pal Singh ·

COMPUTER PE LIKHNE KA MAN NAHIN KAR RAHA THA[3]
THIS IS Q13 NOT QUES 18[2]

1
Grandmaster ·

hey dude try out this one

∫tan-1(six x) dx

39
Dr.House ·

@grandmaster, initially let tant=sinx and

then I=∫ sec2t/√(1-tan2t) dt

now take tant=y

I= ∫ 1/√(1-y2)
=sin-1 y

=sin-1(sin x)

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