Messages posted by akshay_4691

There is NRI quota, admissions based on Class 12 marks.

$\int dx/(1-x^2)$

Put x= $\sin(t)$

so that $dx = \cos(t)dt$

so we have $\int dx/[1-x^2] = \int dt\cos(t)/[1-\sin^2(t)]$

= $\int dt\cos(t)/\cos^2(t) = \int\sec(t)dt$

= $\log(\sec(t)+ \tan(t))$

Resubstitute,

$\sin(t) = x$

i.e. $\sec(t) = \sqrt{1-x^2}$

$\tan(t) = x/[\sqrt{1-x^2}]$

So the integral is = $=\log([x+1]/\sqrt{1-x^2})$

$=\log(\sqrt{1+x/1-x})$

$\int t^2dt/[1-t^2] = \int [t^2-1]dt/[1-t^2] + \int dt /[1-t^2]$

(add and subtract 1)

= $\int -dt + \int dt/[1-t^2]$

= $-t + \log(\sqrt{(1+t)/(1-t)} )$

Apply the condition for Toppling,

The Tension tends to create a clockwise torque about the COM. This causes the normal reaction to shift to the right , and in the limiting case, torque due the normal reaction = 0,when it passes through P

Now, torque due to tension = Th (about P)

Torque due to mg = Mgh/2 (about P )

Now, in the limiting case when the body is just about to topple,

Mgh/2 = Th

Mg/2 =T

Now, Consider the acceleration of block m , a

T-mg= ma

But, T is also equal to Ma , because in the absence of any friction, it would have been T=Ma

Therefore a=g/2

i.e. T-mg = mg/2

T = 3mg/2

since T=Mg/2

Mg=3mg

m/M = 1/3

Observe that x=1 is a root of this equation since ab-ac+bc-ab+ac-bc = 0

if the equation has equal roots, then c(a-b)/a(b-c) = 1

c(a-b) = ab-ac

ac-bc = ab-ac

2ac = ab+bc

b=2ac/(a+c)

i.e. a,b,c are HP

For the question 7, I got a much simpler method.

Put cosx = t , you have,

Integrate by parts,

Take 1 as the function u, and the log term as function v,

$\int \log\sqrt{1-t^2}dt = t\log\sqrt{1-t^2} - \int [\frac{\mathrm{d}\log\sqrt{1-t^2} }{\mathrm{d} x}\int dt]dt$

(differentiation is with respect to "t" and not x )

$\int \log\sqrt{1-t^2}dt = t\log\sqrt{1-t^2} + \int t^2dt/[1-t^2]$

$\int t^2dt/[1-t^2] = \int -dt + \int dt/(1-t^2)$

$\int t^2dt/[1-t^2] = -t + \log(\sqrt{1-t/1+t})$ (theres a typo here, The signs are interchanged within the log term it's actually sqrt(1+t/1-t))

So $\int\log\sqrt{[1-t^2]}dt= -t + \log(\sqrt{1-t/1+t}) + t\log(\sqrt{1+t})+t\log(\sqrt{1-t})$

$\int\log\sqrt{[1-t^2]}dt= -t + (t+1)\log(\sqrt{1+t})+(t-1)\log(\sqrt{1-t})$

Limits from 0 to 1,

$\lim_{x\rightarrow1}(t-1)\log(\sqrt{1-t}) = \lim_{x\rightarrow1}(\log(\sqrt{1-t}/(1/[1-t]))$ (This is a $0.\infty$ form converted into 0/0. Evaluate using L'Hospital Rule)

You will see that this value comes out to be zero

Hence the upper limit value comes out to be $-1 + (1+1)\log(\sqrt{1+1})+ \lim_{x\to 1}(t-1)\log(\sqrt{1-t}) = -1 + 2\log(\sqrt{2}) + 0$

Lower limit t=0 = 0

Hence the value of the integral

$\int_{0}^{\pi/2}\sin(x)\log\sin(x)dx = \int_{0}^{1}\log(\sqrt{1-t^2})dt = \log(2/e)$

$\log$ denotes logarithm to the base "e".

$\int_{0}^{\pi/2}(log\sin(x))\sin(x)dx = I$

INTEGRATE BY PARTS, EVAULATE THE INDEFINITE INTEGRAL FIRST

$\int \log\sin(x)\sin(x)dx = \log\sin(x)\int \sin(x)dx - \int [\frac{\mathrm{d} \log\sin(x)}{\mathrm{d} x}\int \sin(x)dx]dx$

$= -\log\sin(x)\cos(x) + \int [\cos^2(x)/\sin(x)]dx$

$= -\log\sin(x)\cos(x) + \int [\csc(x) - \sin(x)]dx$

$= -\log\sin(x)\cos(x) + \int \csc(x)dx - \int\sin(x)dx$

$= -\log\sin(x)\cos(x) + \log\tan(x/2) + \cos(x)dx$ (TYPO HERE, IGNORE 'dx')

NOW APPLY THE LIMITS,

$-\log\sin(\pi/2)\cos(\pi/2) + \log\tan(\pi/2/2) + \cos(\pi/2) = 0$ .........(1)

NOW FOR THE LOWER LIMIT, AS THE LOG FUNCTIONS BECOME INFINITE AT X=0, WE HAVE TO CONSIDER THE LIMITING VALUE

SO, LOWER LIMIT

$= \lim_{x\rightarrow0}\log(\tan(x/2)/\sin^{\cos(x)}(x)) + \cos(0)$

$= \log(\lim_{x\rightarrow0}((\tan(x/2)/\(\sin^{\cos(x)}(x)))+ \cos(0)$

CONSIDER

$\lim_{x\rightarrow0}((\tan(x/2)/\(\sin^{\cos(x)}(x))$

EVALUATE THIS LIMIT BY L'HOSPITAL RULE,

$\lim_{x\rightarrow0}((\tan(x/2)/\(\sin^{\cos(x)}(x)) =-\log(2)$

SO THE LOWER LIMIT BECOMES

$=-\log(2)+ 1 = -\log(2) + \log(e) = \log(e/2)$ ................(2)

NOW ,

$I = (1)-(2) = -\log(e/2) = \log(2/e)$

http://integrals.wolfram.com/index.jsp?expr=e%5Ex%28tan%5Bx%5D%29&random=false

Don't think this is integrable by standard methods.

Net retardation of the particle = retardation due to friction + retardation to component of gravity along the inclined plane.

Component of gravity along incline = mgsin(theta)

So retardation due to gravity = gsin(theta)

Now, we know that the retardation due to friction is ugcos(theta) where u is the co-efficient of friction.....(1)

By definition the angle of repose R (angle of friction) is such that

tan(R) = u

Since it's given that theta is the angle of repose, tan(theta) = u ......(2)

Using (1) and (2), we have retardation due to friction = g(sinR)(cosR)/(cosR)

ie. Retardation due to friction = gsinR.........(3)

But theta = R and retardation due to gravity = gsin(R) ....(4)

So from (3) and (4), net retardation = 2gsin(R) = 2gsin(θ)

$\int_{0}^{\pi/2}\log\sin(x)dx + \int_{\pi/2}^{\pi}\log\sin(x)dx = \int_{0}^{\pi/2}\log\sin(x)dx + \int_{\pi/2-\pi/2}^{\pi-\pi/2}\log\sin(x+\pi/2)dx$

$= \int_{0}^{\pi/2}\log\sin(x)dx + \int_{0}^{\pi/2}\log\cos(x)dx = \int_{0}^{\pi/2}\log\sin(x)dx + \int_{0}^{\pi/2}\log\sin(x)dx$

$= 2\int_{0}^{\pi/2}\log\sin(x)dx$

BUT

$I1 = (0.5)2\int_{0}^{\pi/2}\log\sin(x)dx$

Hence,

$I1 = \int_{0}^{\pi/2}\log\sin(x)dx = I$

This obviously follows from ,

$I = \int_{0}^{1}\log(x)dx/\sqrt{(1-x^2)}$

Let $\sin^{-1}(x) = t$

so $I = \int_{0}^{\pi/2}\log\sin(t)dt = \int_{0}^{\pi/2}\log\sin(x)dx$

$I = \int_{0}^{\pi/2}\log \sin(x)dx$ I =

$I = \int_{0}^{\pi/2}\log \cos(x)dx$ (integral property)

2I = $\int_{0}^{\pi/2}\log( \sin(x)cos(x))dx$

since 2sinxcosx =sin2x,

$\int_{0}^{\pi/2}\log( \sin(2x)/2)dx$ = 2I

$2I = \int_{0}^{\pi/2}\log( \sin(2x))dx-\int_{0}^{\pi/2}\log(2)dx$

Consider,

$I1 = \int_{0}^{\pi/2}\log(\sin(2x))dx$

Put 2x = t

So you have

$I1 = 0.5\int_{0}^{\pi}\log(\sin(t))dt$

Use property of integrals,

$I1 = 0.5\int_{0}^{\pi/2}\log(\sin(t))dt +0.5\int_{\pi/2}^{\pi}\ln(\sin(t))dt$

Now you can easily prove that I1 = I

hence I = -(pi/2) (log2)

=(pi/2)(log(1/2))

Integrate by parts

Integral = (logx)²INTEGRALdx - INTEGRAL (2logxdx)

= x(logx)² - 2x(logx-1)

T_r = r(r+1)(r+2)(r+3)

Consider a sequence V_r such that,

V_r = r(r+1)(r+2)(r+3)(r+4)

Therefore, V_r-1= (r-1)r(r+1)(r+2)(r+3)

V_r - V_r-1 = r(r+1)(r+2)(r+3)(-r+1 r+4)

= r(r+1)(r+2)(r+3)(5)

I.e. V_r - V_r-1 = 5T_r

So V₁-V₀ = 5T₁

V₂-V₁ = 5T₂

V₃-V₂ = 5T₃ and so on till, V_N - V_N-1 = 5T_N

Adding all these equations gives, T₁+T₂+.....+T_N = (V_N-V₀)/5

Thus Sum = {N(N+1)(N+2)(N+3)(N+4)}0.2

sinx/sin5x = sinx/ {sin3xcos2x+ sin2xcos3x}

Therefore, Denominator = (3-4sin²x)(cos2x) + 2cos3xcosx

= (3-4sin²x)(cos2x) + cos4x+cos2x

= 4cos2x+cos4x + - 4sin²x(cos2x)

= 4cos(2x) + 2cos²2x -1 -4(1-cos²x)(cos2x)

= 4cos²xcos2x-1 +2cos²2x

= 2(1+cos2x)cos2x-1 +2cos²2x

= 2cos2x + 4cos²2x -1

= 4cos²2x + 2cos2x -1

= (cos2x-t1)(cos2x-t2)

i.e. = sinx/sin5x = 1/(cos2x-t1)(cos2x-t2)

t1 = {-1+sqrt(5)}/4

t2 = {-1-sqrt(5)}/4

Hence solve using partial fractions and substituting cos2x = (1-tan²x)/(1+tan²x)

5)

Differentiate using Newton Leibnitz theorem,

f'(x) = 2xcos²x/(1+sin²(x))

f'(pi) = 2(pi)

6) (Not able to post using goiit equation editor so i'll just give you the idea)

Let sin(t) = x

Then t = sin^-1(x)

i.e. dt = dx/{1-x²}^1/2

Therefore the integral reduces to INTEGRAL (0 to pi/2) {log(sint)dt}

Let I = INTEGRAL (0 to pi/2) log(sint)dt

i.e. I = INTEGRAL (0 to pi/2) log(cost)dt

so 2I = INTEGRAL (0 to pi/2) log(sint.cost)dt

= INTEGRAL (0 to pi/2) log(sin2t/2)dt

= INTEGRAL (0 to pi/2)logsin2tdt - INTEGRAL(0 to pi/2)log2dt

But , INTEGRAL (0 to pi/2) log(sin2t)dt = INTEGRAL (0 to pi)(0.5sinzdz) = I

i.e. I = -(pi/2)log(2)

8) Is easy. Use the property INTEGRAL(0 to a) f(x)dx = INTEGRAL (0 to a)f(a-x)dx

So the integral reduces to I = INTEGRAL (0 to pi)0.5(pi)dx/{1+cos²x}

Now, replace cosx = 1/secx. However this makes the function discontinuous at x=(pi/2)

So split the integral as INTEGRAL (0 to pi/2) + INTEGRAL (pi/2 to pi)

Now replace cosx = 1/secx, put tanx = t and proceed.

Pretty straightforward, actually.

Let the mass of the man be "m"

So the mass of the boy becomes m/2

If the speed of the runner be "v", his Kinetic energy = 0.5mv²

If the speed of the boy be v1, his KE = 0.25m(v1)²

Since the runner's KE is half the boy's ,

0.5mv² = 0.125 m(v1)²

i.e. (v1)² = 4v²

ie. v1 = 2v.....(1)

Now when the runner's speed increases by 1 unit, his KE becomes equal to that of the boy's,

i.e. 0.5m(v+1)² = 0.25m(v1)²

Since v1 =2v ,

0.5(v+1)² = 0.25(2v)²

i.e. 2(v+1)² = (2v)²

so sqrt(2)(v+1) = 2v

so v(2-sqrt(2) = sqrt(2)

v = 1/(sqrt(2)-1) unit

Angular velocity after time t seconds = 4t rad/s

i.e. tangential velocity v = 4rt m/s where r is the radius of the circle .

So, the tangential acceleration = 4r m/s²

now, v²/r = 4r

i.e. 16rt²= 4r

t = 0.5 second

Consider the cylinder. Let us assume that the friction force exerted by the plank on the cylinder be f1 and let the friction force exerted by the ground on the cylinder be f2.

We choose the directions arbitrarily. Let f1 be in the backward direction and f2 be in the forward direction.

Consider the free body diagram of the plank m1 .

We have F-f2 = m1(a1)..........(1) (Where a1 is the acceleration of the plank.

Now, considering the linear acceleration of the cylinder,

f2-f1 = m2(a2) ..........(2)

As the cylinder also rotates, and assuming that this is a case of accelerated rolling, we have,

(f1+f2)R = I(k)........(3) where I= MI of cylinder about Centre of Mass and k is its angular acceleration.

Now, since it is a case of accelerated rolling, there is no relative acceleration between any two surfaces in contactand k = (a2)/R ...........(4)

further, 2a2 = a1 ...............(5)

and I = (m2)R²/2 ..............(6)

f2+f1 = (m2)(a2)/2 ...........(7)

Solve these 7 equations for f1, f2, a1, a2.