Calculus - integration siikho

39

Dr.House ·2009-03-03 10:10:34

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

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39

Dr.House ·2009-03-03 18:10:45

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}

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39

Dr.House ·2009-03-03 18:54:35

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 xâ‰¥0

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

38]Let f(x) be the function defined for xâ‰¥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)

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39

Dr.House ·2009-03-03 21:40:09

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 Î± 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

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13

Ð”Ð²Ò¥Ñ—ÑuÏ now in medical c ·2009-03-03 22:58:16

5]âˆ«x²dx/(cosx+xsinx)²

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

and on simplification

=xsecxâˆ«d(xsinx+cosx)/(cosx+xsinx)²-âˆ«sec²x(xsinx+cosx)dx/xsinx+cosx

=-xsecx/xsinx+cosx + tanx + C

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106

Asish Mahapatra ·2009-03-04 03:52:26

6) f(cosx) = cos²x + lcosxl
= cos²x + cosx .... x Îµ [0,Ï€/2]
= cos²x - cosx .... x Îµ [Ï€/2,Ï€]
so. âˆ«f(cosx) from 0 to pie/2 is âˆ«(1+cos2x)/2dx + âˆ«cosxdx
= [x + sin2x/2]/2 + sinx
= Ï€/4 + 1

âˆ«f(cosx) from pie/2 to pie is âˆ«(1+cos2x)/2dx - âˆ«cosxdx
= [x + sin2x/2]/2 - sinx
= Ï€/2 - (Ï€/4-1)
= Ï€/4 + 1

so âˆ«f(cosx)dx from 0 to pie = Ï€/2 + 2 ...... (i)

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

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

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106

Asish Mahapatra ·2009-03-04 04:03:21

(3) âˆ«sin⁴dx = âˆ«(1-cos2x)²dx/4 = âˆ«(1-2cos2x+cos²2x)dx/4
= âˆ«[1-2cos2x+ (1+cos4x)/2]dx/4
= âˆ«(2-4cos2x+1+cos4x)dx/8
= âˆ«(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(aâ†’âˆž)[âˆ«sin⁴dx]/a = lim(aâ†’âˆž)[4-8sin2a/a + sin4a/a]/32
= 1/8............

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1

Akand ·2009-03-06 04:36:43

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

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1

Akand ·2009-03-06 04:42:44

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

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11

Subash ·2009-03-06 04:42:45

24

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

then take sinx/2 =t

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1

Akand ·2009-03-06 04:46:21

26... I=âˆ«1.sin^-1xdx
ILATE........
I= xsin^-1x-âˆ«x/âˆš1-x²dx
let 1-x²=t²
so -2xdx=2tdt
I=xsin^-1x+âˆ«tdt/t
I=xsin^-1x+âˆš1-x²+c

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1

Akand ·2009-03-06 05:05:52

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

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1

Akand ·2009-03-06 05:14:44

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

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1

Akand ·2009-03-06 05:20:40

23....
y=âˆ«a^-x/2dx
logy=-loga/2âˆ«xdx........I dunno whether i can do this..
logy=-x²loga/4+c
logy=loga^-x²/4+c
y=e^{loga^-x²/4+c}
y=a^-x²/4.e^c

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1

Akand ·2009-03-06 05:28:31

27.
I=y²âˆšy²-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)²âˆšy²-a²/5+ (y²-a²)âˆšy²-a²/3+c

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1

Akand ·2009-03-06 05:39:34

29.
SORRY MADE A MISTAKE.

f(cosx)=cos⁴x+cosx ........for 0 to âˆ©/2
f(cosx)=cos⁴x-cosx..........forâˆ©/2 to âˆ©
₀âˆ«^âˆ©=₀âˆ«^âˆ©/2+_âˆ©/2âˆ«^âˆ©
solving this.........
..........
I₁=3âˆ©/8
and I₂=3âˆ©/16
I₁/I₂=2

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11

virang1 Jhaveri ·2009-03-06 05:43:01

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

Thx Bhaiyya

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11

virang1 Jhaveri ·2009-03-06 06:04:02

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

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1

Akand ·2009-03-06 06:54:38

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........

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1

Akand ·2009-03-06 07:37:07

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

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1

Akand ·2009-03-06 07:39:35

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

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62

Lokesh Verma ·2009-03-06 07:52:03

2t²-t+1

t²-t/2+1/2

(t-1/4)²+ .....

now is this done?

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1

Akand ·2009-03-06 07:53:19

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

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1

Akand ·2009-03-06 07:57:11

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