Q.469. Indefinite integration.

Question 469.
Integrate cosec^5 (5x) dx.

Answer 469.
Let 5x = u
=> 5 dx = du
=> Integrand
= (1/5) ∫ cosec^5 u du ... ( 1 )

∫ cosec^3 u du
= ∫ cosecu * cosec^2 u du
Using integration by parts
= cosecu ∫ cosec^2 u du - ∫ [d/du(cosecu) ∫ cosec^2u du] du
= - cosecu cotu - ∫ cosecu cot^2 u du
= - cosecu cotu - ∫ cosecu (cosec^2 u - 1) du
= - cosecu cotu - ∫ cosec^3 u du + ∫ cosecu du
=>
2 ∫ cosec^3 u du
= - cosecu cotu + ln ltan(u/2)l + 2c
=> ∫ cosec^3 u du
= - (1/2) cosecu cotu + (1/2) ln ltan(u/2)l + c ... ( 2 )

Now, ∫ cosec^5 u du
= ∫ cosec^3 u * cosec^2 u du
Integrating by parts,
= cosec^3 u ∫ cosec^2 u du - ∫ [d/du(cosec^3 u) ∫ cosecu du] du
= - cosec^3 u * cotu - ∫ 3 cosec^3 u * cot^2u du
= - cosec^3 u * cotu - 3 ∫ cosec^3u (cosec^2u - 1) du
= - cosec^3 u * cotu - 3 ∫ cosec^5 u du + 3 ∫ cosec^3 u du
=> 4 ∫ cosec^5 u du
= - cosec^3 u * cotu + 3 ∫ cosec^3 u du

[Plugging the value of ∫ cosec^3 u du from ( 2 ) above]
= - cosec^3 u * cotu + 3 [ - (1/2) cosecu cotu + (1/2) ln ltan(u/2)l ] + 4c
=> ∫ cosec^5 u du
= - (1/4) cosec^3 u * cotu - (3/8) cosecu cotu + (3/8) ln ltan(u/2)l ] + c

[Plugging in ( 1 ) above]
=> Integrand
= - (1/20) cosec^3 (5x) * cot(5x) - (3/40) cosec(5x) cot(5x)
   + (3/40) ln ltan(5x/2)l + c'. [c' = c/5]

Confirmation that the above answer is correct as verified by Wolfram Alpha:
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Q.462. Integration.

Question 462.
Find  (x^3) / [(x^2) + 4x + 8] dx.

Answer 462.
Note that the numerator is a polynomial of degree 3 and the denominator is of degree 2.
Hence, the first step is to perform a division and express the function in the form of a quotient + remainder/denominator so that the remainder is a polynomial of degree less than the denominator. Integration follows thereafter. Thus,
x^3
= x (x^2 + 4x + 8) - 4x^2 - 8x
= x (x^2 + 4x + 8) - 4(x^2 + 4x + 8) + 8x + 32

=> x^3/(x^2 + 4x + 8)
= (x - 4) + 8 (x + 4) / (x^2 + 4x + 8)

=> Integral
= ∫ (x - 4) dx + 4 ∫ (2x + 4 + 4) / (x^2 + 8x + 8) dx
= x^2/2 - 4x + 4 ∫ (d(x^2 + 8x + 8) / (x^2 + 8x + 8) + 16 ∫ dx / [(x + 2)^2 + 2^2]
= x^2/2 - 4x + 4 log(x^2 + 8x + 8) + 8 tan^-1 [(x + 2)/2] + c.

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Q.454. Application of differentiation.

Question 454.
At noon, ship A is 100 km west of ship B. Ship A is sailing east at 35km/hr and ship B is sailing north at 25 km/hr. How fast is the distance between the ships changing at 4.00 p.m.


Answer 454.
Let the position of ship B at noon be at the origin
=> ship A has x-coordinate = - 100 km
and B has x-coordinate = 0 km
At 4.00 p.m., ship A is at x = - 100 + 4 * 35 = 40 km
and B is at x = 0 and y = 4 * 25 = 100 km
Distance between them at 4.00 p.m.
= √[(100)^2 + (40)^2] = 20√(29) km

Distance between the ships,
s^2 = x^2 + y^2
=> 2s ds/dt = 2x dx/dt + 2y dy/dt
=> rate of change of distance between the ships, ds/dt
= (x dx/dt + y dy/dt) / s
= [40 * 35 + 100 * 25] / [20√(29)] km/hr
= (1400 + 2500) / [20√(29)] km/hr
= 195/√(29) km/hr
≈ 36.21 km/hr.

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Q.448. Rate of change (Application of Differentiation)

Question 448.
At what rate is the viewing angle changing when the observer is 50 feet from the tunnel if they are traveling at 6 feet per second? The picture is as under.

Answer 448.
θ
= arctan(25/50) - arctan(15/50)
= arctan(1/2 - 3/10) / (1 + 3/20)
= arctan (4/23)

If the observer were x ft. away,
θ
= arctan(25/x - 15/x)/(1 + 375/x^2)
= arctan [10x/(x^2 + 375)]
=> tanθ = 10x/(x^2 + 375)
=> sec^2 θ dθ/dt = [(x^2 + 375) * 10 - 10x * 2x]/(x^2 + 375)^2 dx/dt ... ( 1 )

sec^2 θ = 1 + tan^2 θ = 1 + 4/23 = 27/23
Plugging sec^2 θ = 27/23, dx/dt = 6 and x = 50 in ( 1 )
=> dθ/dt
= [(2500+375)*10 - 20*2500)] * 6/[(27/23) * (2500+375)^2] rad/sec
= (28750 - 50000) * 6 / [(27/23) * (8265625) rad/sec
= - (21250 * 23) * 6 / (27 * 8265625) rad/s
= - 0.01314 rad/s.
[Negative sign indicates that the angle is decreasing.]

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Q.439. Area of the surface of revolution

Question 439.
Find area of the surface of revolution generated by revolving around x axis
one loop of (8y^2) = (x^2)(1-x^2).

Answer 439.
Plugging y = 0
=> x^2 (1 - x^2) = 0
=> x = -1, 0 or 1
=> one loop is from x = 0 to x = 1

8y^2 = x^2 (1 - x^2) = x^2 - x^4
=> 16ydy/dx = 2x - 4x^3
=> ydy/dx = (1/8) (x - 2x^3)

Surface area, S
= 2π ∫ (x = 0 to 1) y √[1 + (dy/dx)^2] dx
= 2π ∫ (x = 0 to 1) √[y^2 + (ydy/dx)^2] dx
= 2π ∫ (x = 0 to 1) √[ (x^2/8)(1 - x^2) + (1/64)(x - 2x^3)^2] dx
= (π/4) ∫ (x = 0 to 1) √(8x^2 - 8x^4 + x^2 - 4x^4 + 4x^6) dx
= (π/4) ∫ (x = 0 to 1) √(9x^2 - 12x^4 + 4x^6) dx
= (π/4) ∫ (x = 0 to 1) (3x - 2x^3) dx
= (π/4) (3x^2/2 - x^4/2) ... (x = 0 to 1)
= π/4.

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Q.435. Differential Equation.

Question 435.
sinx.cosx.dy/dx - y = sinx

Answer 435.
sinx cos x dy/dx - y = sin x
=> tanx dy/dx - y sec^2x = sec x tan x (dividing by cos^2 x)
=> (tanx dy/dx - y sec^2x) / tan^2 x = csc x (dividing by tan^2 x)
=> d/dx ( y / tan x ) = csc x
=> d ( y / tan x) = csc x dx

Integrating,
y / tan x = ∫ csc x dx - ln c
=> y / tan x = ln [ l tan (x/2) l / c ]
=> tan (x/2) / c = e^(y/tan x)
=> tan(x/2) = c*e^(y/tan x)

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Q.434. Application of integration to find the volume of a solid.

Question 434.
A solid has an isosceles triangular base of side length 10, 13, 13 inches. Cross sections parallel to one of the two equal length sides are equilateral triangles. Find the volume of the solid with the exact value.

Answer 434.
Let in ΔABC,
AB = AC = 13 and BC = 10
Consider a small length DD' on BC and EE' on AC such that DE and D'E' are parallel to AB
Let BD = x and DD' = dx
=> DE = x * (13/10)
=> Area of equilateral triangle of length DE
= (1/2) (13x/10)^2 sin60° = x * 13√3/40
and volume of the thin slice of width DD'
dV = (13√3/40) ∫ x dx ... (x=0 to 10)
=> V = (13√3/80) x^2 ... (x=0 to 10)
=> V = 65√3/4 cubic units.

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Q.430. Homogeneous differential equation.

Question 430.
Solve and get the general solution of the differential equation:
xy dx - (x^2 + 3y^2) dy = 0.

Answer 430.
xy dx - (x^2 + 3y^2) dy = 0
=> dy/dx = xy/(x^2 + 3y^2)
This is a homogeneous differential eqn. of the first order.
Let y = vx
=> dy/dx = v + x dv/dx
=> v + x dv/dx = vx^2 / (x^2 + 3v^2x^2)
=> x dv/dx = v/(1 + 3v^2) - v
=> x dv/dx = (v - v - 3v^3) / (1 + 3v^2)
=> x dv/dx = - 3v^3 / (1 + 3v^2)
=> (1 + 3v^2) / 3v^3 dv + dx/x = 0
=> (1/3) ∫ dv/v^3 + ∫ dv/v + ∫ dx/x = 0
=> - (1/6v^2) + lnv + lnx = lnc
=> - (x^2/6y^2) + lny = lnc
=> ln(y/c) = x^2/6y^2
=> y = c e^(x^2/6y^2).

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Q.426. Differential Equation

Question 426.
Find the particular solution of the differential equation,
dy/dt = 0.5-t + 2y,  y(0) = 2.

Answer 426.
dy/dt = 0.5 - t + 2y
=> dy/dt - 2y = 0.5 - t
=> e^-2t dy/dt - 2y e^-2t = (0.5 - t)e^(-2t)
=> d(ye^-2t) = (0.5 - t)e^(-2t) dt

Integrating,
y e^(-2t)
= ∫ (0.5 - t) e^(-2t) dt
= (0.5) ∫ e^(-2t) dt - ∫ t e^(-2t) dt
= - (1/4) e^(-2t) - [ t ∫ e^(-2t) dt - ∫ [d/dt(t) ∫ e^(-2t) dt] dt]
= - (1/4) e^(-2t) - [- (1/2) te^(-2t) + (1/2) ∫ e^(-2t) dt]
= - (1/4) e^(-2t) - [- (1/2) te^(-2t) - (1/4) e^(-2t)] + c/2
= (1/2) te^(-2t) + c/2
=> ye^(-2t) = (1/2) te^(-2t) + c/2
=> 2y = t + ce^(2t)

Plugging t = 0 and y = 2
=> 4 = c
=> particular solution is
2y = t + 4e^(2t)
=> y = t/2 + 2e^(2t).

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Q.409. Formation of differential equation representing family of circles.

Question 409.
Prove that the differential equation of the family of circles, all touching x-axis is given by
(1 + y'^2)^3 = [yy" + (1 + y'^2)]^2.

Answer 409.
Let (h, k) be the centers of the circles with h and k as arbitrary constants.
As the circles touch the x-axis, radius, r = l k l
=> general equation of the family of circles is
(x - h)^2 + (y - k)^2 = k^2 ... ( 1 )


Differentiating w.r.t. x,
=> 2(x - h) + 2(y - k) y' = 0
=> (x - h) = - (y - k)y'
=> (x - h)^2 - (y - k)^2 y'^2 = 0 ... ( 2 )


( 1 ) - ( 2 )
=> (y - k)^2 (1 + y'^2) = k^2 ... ( 3 )

Differentiating w.r.t. x,
2(y - k) y' (1 + y'^2) + (y - k)^2 * 2y'y" = 0

Dividing by 2y'(y - k)
=> (1 + y'^2) + (y - k) y" = 0
=> y - k = - (1 + y'^2)/y" and k = y + (1 + y'^2)/y"
Plugging y - k and k from above in eqn. ( 3 ),
[ - (1 + y'^2)/y"]^2 (1 + y'^2) = [y + (1 + y'^2)/y"]^2
=> (1 + y'^2)^3 = [yy" + (1 + y'^2)]^2

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Q.405. Reduction Formula in Integration.

Question 405.
If I(n) = ∫ (from 0 to pi/2) x sin^n x dx,
prove that I(n) = [ (n-1)/n ] I(n-2) + 1/n^2.

Answer 405.
I(n)
= ∫ (from 0 to π/2) x sin^n x dx
= ∫ (from 0 to pi/2) x sin^(n-1) x sinx dx
= [x sin^(n-1)x (-cosx)] ... (0 to π/2)
    + ∫ (from 0 to π/2) [sin^(n-1) x
    + x (n-1) sin^(n-2) x cosx] cosx dx
= ∫ (from 0 to π/2) [sin^(n-1) x cosx dx
    + (n-1) ∫ (from 0 to π/2) x sin^(n-2) * (1 - sin^2 x) dx
= (1/n) sin^n x ... (x = 0 to π/2) + (n-1) [I(n-2) - In]
=> (1 + n - 1) In = 1/n + (n-1) I(n-2)
=> In = 1/n^2 + [(n-1)/n] I(n-2).

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Q.403. Reduction formula in integration

Question 403.
If I(n) = ∫ (from 0 to pi/2) cos^n x sin nx dx,
prove that 2I(n+1) = I(n) + 1/(n+1).

Answer 403.
I(n+1)
= ∫ (x=0 to π/2) cos^(n+1)x sin(n+1) x dx
Using integration by parts
= [cos^(n+1)x ∫ sin(n+1)x dx] ... (x=0 to π/2)
          - ∫ (x=0 to π/2) [d/dx (cos^(n+1)x) ∫ sin(n+1)x dx] dx
= - 1/(n+1) [cos^(n+1)x cos(n+1)x] ... (x=0 to π/2)
   - ∫ (x=0 to π/2) [1/(n+1)cos^n x * (-sinx) * 1/(n+1) * (-cos(n+1)x) dx
= 1/(n+1) - ∫ (x=0 to π/2) cos^n x cos(n+1)x sinx dx ... ( 1 )

sin(nx) = sin[(n+1)x - x] = sin(n+1)x cosx - cos(n+1)x sinx
=> cos(n+1)x sinx = sin(n+1)x cosx - sin(nx)

Plugging this result in ( 1 ),
I(n+1)
= 1/(n+1) - ∫ (x=0 to π/2) cos^n x *[sin(n+1)x cosx - sin(nx)] dx
= 1/(n+1) - ∫ (x=0 to π/2) cos^(n+1)x sin(n+1) x dx + ∫ (x=0 to π/2) cos^n x sin(nx) dx
= 1/(n+1) - I(n+1) + I(n)
=>
2I(n+1) = I(n) + 1/(n+1).

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Q.400. Reduction formula in integration

Question 400.
Obtain a reduction formula for :
I(n) = ∫ (from 0 to pi/2) x^n sin x dx .....where n > 1
and hence evaluate I(3).
=> I(n) + n(n-1) I(n-2) = n (π/2)^(n-1)

Answer 400.
Integrating by parts,
I(n)
= [-x^n cosx] (x=0 to π/2) + n ∫x^(n-1) cosx dx ... (x=0 to π/2)
= 0 + n [ {x^(n-1) sinx} (x=0 to π/2) - (n-1) ∫ x^(n-2) sinx dx ] ... (x=0 to π/2)
= n (π/2)^(n-1) - n(n-1) I(n-2)

I(3)
= 3(π/2)^2 - 6 I(1)
= 3(π/2)^2 - 6 ∫ (from 0 to π/2) x sin x dx ... ( 1 )

∫ (from 0 to π/2) x sin x dx
= - x cosx + ∫ cosx dx ... (x=0 to π/2)
= - x cosx + sinx ... (x=0 to π/2)
= 0 + 1

Plugging in ( 1 ),
I(3) = 3(π/2)^2 - 6.

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Q.392. Question on Lagrange's Mean value theorem.

Question 392.
Prove that 0 < [ln (1+x) ]^-1 - x^-1 < 1 for all x > 0 using Mean Value Theorem.

Answer 392.
Let f (x) = ln (1 + x)
=> by MVT for [0, x],
[ln (1 + x) - ln1] / (x - 0) = 1/(1 + c), 0 < c < x
=> c = x/ln(1 + x) - 1
=> 0 < c = x/ln(1 + x) - 1 < x
=> 0 < x/ln(1 + x) - 1 < x
=> 0 < [x - ln(1 + x)] / xln(1 + x) < 1
=> 0 < [ln(1 + x)]^-1 - x^-1 < 1.

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Q.390. Application of Mean Value Theorem

Question 390.
Using Lagrange's mean value theorem proove that :
x < sin ^-1 x < x / [square root of (1-x^2) ] for 0 < x < 1.

Answer 390.
Let f (x) = sin^-1 x
f (x) is continuous in [0, 1] and differentiable in (0, 1)
Hence, MVT is applicable and
f (a + h) = f (a) + h f '(a + θh), 0 < θ < 1
Taking a = 0, h = x and f '(x) = 1/√(1 - x^2)
=> sin^-1 x = x / √(1 - θ^2x^2)
=> √(1 - θ^2x^2) = x/sin^-1 x ... ( 1 )

0 < θ < 1
=> 0 < θ^2 x^2 < x^2
=> 0 > - θ^2 x^2 > - x^2
=> - x^2 < - θ^2 x^2 < 0
=> 1 - x^2 < 1 - θ^2 x^2 < 1
=> √(1 - x^2) < √(1 - θ^2 x^2) < 1

Plugging √(1 - θ^2x^2) = x/sin^-1 x from ( 1 ),
=> √(1 - x^2) < x/sin^-1 x < 1
=> 1/√(1 - x^2) > sin^-1x / x > 1
=> 1 < sin^-1 x / x < 1/√(1 - x^2)
=> x < sin^-1 x < x/√(1 - x^2).

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Q.388. Definite integration.

Question 388.
Evaluate ∫ (x = 0 to π) (a^2 cos^2 x + b^2 sin^2 x) dx

Answer 388.
Using the formula
∫ (x = 0 to π) f (x) dx = 2 ∫ (x = 0 to π/2) f (x) dx, if f (π - x) = f (x),
=> ∫ (x = 0 to π) (a^2 cos^2 x + b^2 sin^2 x) dx
= 2 ∫ (x = 0 to π/2) (a^2 cos^2 x + b^2 sin^2 x) dx ... ( 1 )
= 2 ∫ (x = 0 to π/2) [a^2 cos^2 (π/2 - x) + b^2 sin^2 (π/2 - x)] dx
.........................Using ∫ (x = 0 to a) f (x) dx = ∫ (x = 0 to a) f (a - x) dx]
= 2 ∫ (x = 0 to π/2) (a^2 sin^2 x + b^2 cos^2 x) dx ... ( 2 )

Adding ( 1 ) and ( 2 ),
2 ∫ (x = 0 to π) (a^2 cos^2 x + b^2 sin^2 x) dx
= 2 ∫ (x = 0 to π/2) [a^2 (cos^2 x + sin^2 x) + b^2 (sin^2 x + cos^2 x)] dx
=> ∫ (x = 0 to π) (a^2 cos^2 x + b^2 sin^2 x) dx
= ∫ (x = 0 to π/2) (a^2 + b^2) dx
= (a^2 + b^2) [x] ... (x = 0 to π/2)
= (π/2) (a^2 + b^2).

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Q.383. Maximizing area of cylinder inscribed in a sphere.

Question 383.
What is the largest surface area of a right circular cylinder inscribed in a sphere of radius r ?

Answer 383.
Refer to the figure shown below.
Let radius of the cylinder, R = rcosA
and height of the cylinder, H = 2rsinA
=> surface area of the cylinder,
S = 2πR^2 + 2πRH

=> S = 2πr^2 (cos^2 A + 2 sinA cosA)

=> S = πr^2 (1 + cos2A + 2sin2A)

For S to be maximum, dS/dA = 0 and d^2S/dA^2 < 0

dS/dA = 0

=> πr^2 (- 2sin2A + 4cos2A) = 0

=> tan2A = 2 => sin2A = 2/√5 and cos2A = 1/√5

d^2A/dA^2 = πr^2 (-4cos2A - 8sin2A) < 0 for acute A
=> maximum surface area,
S(max)
= πr^2 (1 + 1/√5 + 4/√5)

= πr^2 (1 + √5).

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Q.382. Second order implicit differentiation.

Question 382.
Use implicit differentiation to find dy/dx and then d^(2)y/dx^2 of y^2 = 7x^(2) - 6x.

Answer 382.
y^2 = 7x^2 - 6x
=> 2y dy/dx = 14x - 6
=> dy/dx = (7x - 3) / y   ...   ( 1 )

Differentiating ( 1 ) wrt. x,
y d^2y/dx^2 + (dy/dx)^2 = 7
=> d^2y/dx^2
= [7 - (dy/dx)^2] / y
= [7 - (7x - 3)^2 / y^2] / y
= [7y^2 - (7x - 3)^2] / y^3
= [7 * (7x^2 - 6x) - (49x^2 - 42x + 9)] / y^3
= - 9/y^3.

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Q.377. Indefinite integration.

Question 377.
Find ∫ [(1 + sinx) / (1 + cosx)] e^x dx.

Answer 377.
e^x * [(1 + sinx) / (1 + cosx)]
= e^x * [1 + 2sin(x/2) cos(x/2)] / [2cos^2 (x/2)]
= [(1/2) sec^ (x/2) + tan(x/2)] e^x

 Let f (x) = tan(x/2)
=> f '(x) = (1/2) sec^2 (x/2)

 => ∫ e^x [(1+sinx)/(1 + cosx)] dx
= ∫ [(1/2) sec^ (x/2) + tan(x/2)] e^x dx
= ∫ [ f (x) + f '(x) ] e^x dx
= f (x) e^x + c
= tan(x/2) e^x + c.

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Q.369. Integration.

Question 369.
Evaluate  ∫ dx / (x^2 + 2x + 2)^2.

Answer 369.
x^2 + 2x + 2
= (x + 1)^2 + 1

Let x + 1 = tanu
=> (x + 1)^2 + 1 = tan^2 u + 1 = sec^2 u
and dx = sec^2 u du

=> Integral
= ∫ sec^2 u du / sec^4 u
= (1/2) ∫ 2cos^2 u du
= (1/2) ∫ (1 + cos2u) du
= (1/2) u + (1/4)sin2u + c
= (1/2) u + (1/2) tanu / (1 + tan^2 u) + c
= (1/2) arctan (x + 1) + (1/2) (x + 1) / [(x + 1)^2 + 1] + c
= (1/2) arctan (x + 1) + (1/2)(x + 1)/(x^2 + 2x + 2) + c
= (1/2) [arctan(x + 1) + (x + 1)/(x^2 + 2x + 2)] + c.

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