integration siikho

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

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}

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)

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

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

6) f(cosx) = cos²x + lcosxl
= cos²x + cosx .... x ε [0,Ï€/2]
= cos²x - cosx .... x ε [Ï€/2,Ï€]
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,Ï€/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...

(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............

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........

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

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

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=e^loga-x2/4+c
y=a^-x2/4.e^c

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)√y²-a²/3+c

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

Thx Bhaiyya

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

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........

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

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

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........

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

34]
RATIONALISE N THEN SPLIT;

ANS : 4√x - x - ln(1+√x) + c

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

hey dude try out this one

∫tan^-1(six x) dx

@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)