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:
Wolfram Alpha Link

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Q.465. Remainder in a large division.

Question 465.
What is the remainder if 5^2009 + 13^2009 is divided by 18 ?

Answer 465.
Using,
For an odd positive integer n,
x^n + y^n
= (x + y) [x^(n-1) - x^(n-2)y + x^(n-2)y^2 - ... + y^(n-1)]

5^2009 + 13^2009
= (5 + 13) [5^2008 - 5^2007 * 13 + 5^2006 * 13^2 - ... + 13^2008]
= 18 * [5^2008 - 5^2007 * 13 + 5^2006 * 13^2 - ... + 13^2008]
=> 5^2009 + 13^2009 is divisible by 18 with zero remainder.

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Q.464. To find the equation of the in-circle given vertices of a triangle

Question 464.
Find the equation of circle inscribed in the triangle with vertices
(-7,-10),(-7,15) and (5,-1).

Answer 464.
Let A (-7, -10), B (-7, 15) and C (5, -1) be the given vertices of ΔABC

=> a = BC = √[(5+7)^2 + (-1-15)^2] = 20,
b = CA = √[(5+7)^2 + (-1+10)^2] = 15
and c = AB = √(-7+7)^2 + (15+10)^2] = 25

=> coordinates of the in-center are
x = 1/(a+b+c) * (-7a -7b+5c)
= 1/(20+15+25) * (-7*20 - 7*15 + 5*25) = - 2
and y = 1/(a+b+c) * (-10a +15b-c)
= 1/(20+15+25) * (-10*20 +15*15 - 25) = 0
=> incenter = (-2, 0)

Radius of the in-circle
= perpendicular distance from (-2, 0) to the line through AB
= -2 + 7 = 5

=> eqn. of the inscribed circle is
(x + 2)^2 + y^2 = 5^2
=> x^2 + y^2 + 4x - 21 = 0.

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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.457. Trigonometric inequality.

Question 457.
Prove that (sin^3 A)/(sin B) + (cos^3 A)/(cos B) ≥ sec ( A - B ) for all 0 < a,b < π/2.

Answer 457.
(sin^3 A)/(sin B) + (cos^3 A)/(cos B) ≥ sec ( A - B )
<=> (sin^3 A cosB + cos^A sinB) / sinB cosB ≥ 1/cos(A - B)
<=> (sin^3 A * cosB + cos^3 A * sinB) * cos(A - B) ≥ sinB cosB
<=> (sin^2 A * 2sinA cosB + cos^2 A * 2cosA sinB) * cos(A - B) ≥ 2sinB cosB
<=> [sin^2 A {sin(A+B) + sin(A-B)} + cos^2 A {sin(A+B) - sin(A-B)}] * cos(A - B) ≥ sin2B
<=> [(sin^2 A + cos^2 A) sin(A + B) - (cos^2 A - sin^2 A) sin(A - B)] * cos(A - B) ≥ sin2B
<=> sin(A + B) cos(A - B) - cos2A sin(A - B) cos(A - B) ≥ sin2B
<=> 2sin(A + B) cos(A - B) - cos2A * 2sin(A – B) cos(A - B) ≥ 2sin2B
<=> sin2A + sin2B - cos2A * 2sin(A – B) cos(A - B) ≥ 2sin2B
<=> sin2A - sin2B - cos2A * 2sin(A – B) cos(A - B) ≥ 0
<=> 2cos(A + B) sin(A - B) - cos2A * 2sin(A - B) cos(A - B) ≥ 0
<=> sin(A - B) [cos(A + B) - cos2A cos(A - B)] ≥ 0
<=> sin(A - B) [2cos(A + B) - 2cos2A cos(A - B)] ≥ 0
<=> sin(A - B) [2cos(A + B) - cos(3A - B) - cos(A + B)] ≥ 0
<=> sin(A - B) [cos(A + B) - cos(3A - B)] ≥ 0
<=> 2sin2A sin^2 (A - B) ≥ 0
which is true
=> the given inequality is true.

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Q.455. Geometry challenge

Question 455.
A circle (radius = r), and an equilateral triangle (side = 2r), fit perfectly in a square, as shown in the diagram.
What is (length CD) divided by (height of triangle)?

Answer 455.

Trigonometric Proof:

Refer to the figure:

Let O be the center of the circle.
With r = 1, (The required ratio is independent of the value of r.)
vertical side of the square
= 1 + ODcos60° + CDcos30°
= 1 + 1/2 + (√3/2) CD
= 3 + (√3/2) CD ... ( 1 )

Horizontal side of the square
= 1 + ODcos30° + (BC - CDcos60°)
= 1 + √3/2 + 2 - CD/2
= 3 + √3/2 - (1/2) CD ... ( 2 )
Equatting ( 1 ) and ( 2 ),
3/2 + (√3/2) CD = 3 + √3/2 - (1/2) CD
=> (√3 - 1)/2 CD = (√3 - 3)/2
=> CD = √3
and Height of the triangle = 2 cos30° = √3
=> CD/Height of the triangle
= √3 / √3
= 1.
=======================================…
Proof using Co-ordinate Geometry:
Refer to the figure:

Consider the given drawn inverted as above.

The required ratio, being independent of the radius, let r = 1
=> The eqn. of the circle with its center as origin is
x^2 + y^2 = 1 ... ( 1 )

Let the length of side of the square = a
=> B = (a-1, a-1), C = (a-3, a-1) and A = (a-2, a-1-√3)
Slope of AC = - √3
Let the eqn. of the tangent AC be y = - √3x + c
=> c = r √(1 + m^2) = 2
=> eqn. of AC is y = - √3x + 2 ... ( 2 )

Solving eqn. ( 1 ) and ( 2 ) gives
x^2 + (-√3x + 2)^2 = 1
=> 4x^2 - 4√3x + 3 = 0
=> (2x - √3) = 0
=> x-coordinate of D is √3/2
Plugging in eqn. ( 2 ),
y-coordinate is 1/2
=> D = (- √3/2, 1/2)

Plugging coordinates of C in eqn. ( 2 ),
a -1 = - √3 (a - 3) + 2
=> a = (3√3 + 3) / (√3 + 1) = 3
=> C = (0, 2)

CD^2 = (0 + √3/2)^2 + (3/2)^2 = 3
=> CD = √3
and Height of the triangle = 2 cos30° = √3
=> CD/Height of the triangle
= √3 / √3
= 1.

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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.452. Trigonometric inequality.

Question 452.
In any triangle ABC,  prove that : sin A sin B sin C  ≤  3√3/8.

Answer 452.
f = sinA sinB sinC
= sinA sinB sin[π - (A+B)]
= sinA sinB sin(A+B)
= sinA sinB (sinA cosB + cosA sinB)
= sin^2 A sinB cosB + sin^2 B sinA cosA

To find the maximum value,
δf/δA = 0
=> 2sinA cosA sinB cosB + sin^2 B * cos2A = 0
=> sin2A cosB + cos2A sinB = 0
=> sin(2A + B) = 0
=> 2A + B = π ... ( 1 ) [cannot be zero or higher multiple of π for traingle ABC]

Similarly,
δf/δB = 0
=> 2B + A = π ... ( 2 )

Solving ( 1 ) and ( 2 ),
A = B = π/3 => C = π/3
=> sinA sinB sinC = [sin(π/3)]^3 = (3√3)/8
This can be shown to be maximum and not minimum by taking some arbitrary values of A, B and C.
Taking A = π/2, B = C = π/4
=> sinA sinB sinC = 1/2
which shows that (3√3)/8 is the maximum value of sinA sinB sinC
=> sinA sinB sinC ≤ (3√3)/8.
=======================================…

For the benefit of the readers, I reproduce the elegant solution given on page 4-66 of the following link.

Click to open link.

Consider three points P(A, sinA), Q(B, sinB) and R(C, sinC)
on the curve y = sinx such that A + B + C = π
Centroid of ΔPQR = [π/3, (1/3)(sinA + sinB + sinC)]
Therefore, centroid lies on x = π/3 and is inside the triangle,
=> (1/3) (sinA + sinB + sinC) ≤ sin(π/3) = √3/2
=> (sinA sinB sinC)^(1/3) ≤ √3/2 ... [ GM ≤ AM ]
=> sinA sinB sinC ≤ (3√3)/8.

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Q.451. Challenging area problem of geometry.

Question 451.
Equilateral triangle PQR is on square ABCD, as at the picture:
For given areas of right triangles (22 and 23) find unknown trapezoidal area, exact value.

Answer 451.
Let ∠ DRP = x
=> ∠ CRQ = 120° - x
Let a = length of the side of the equilateral triangle
=> (1/2) acosx * asinx = 22
=> a^2/4 sin2x = 22 ... ( 1 )
Similarly,
a^2/4 sin(240° - 2x) = 23 ... ( 2 )

Taking ratio ( 2 ) to ( 1 ),
sin(240° - 2x) / sin2x = 23/22
=> 22 * [- (√3/2) cos2x + (1/2) sin2x] = 23 * sin2x
=> - 11√3 cos2x = 12sin2x
=> tan2x = - (11√3)/12
=> sin2x = (11√3)/13√3 = 11/13
and cos2x = sin2x/tan2x = - 12/(13√3)

Plugging sin2x in ( 1 ),
a^2 = 88 / (11/13) = 104

Area of triangle = (√3/4) a^2 = 26√3

Length of side of the square
= acosx + acos(120° - x)
= acosx - (a/2)cosx + (a√3/2) sinx
= (a/2) cosx + (a√3/2) sinx

Area of the square
= [(a/2) cosx + (a√3/2) sinx]^2
= a^2 [(1/4) cos^2 x + (3/4) sin^2 x + √3 sinx cosx]
= (104) * [(1/8)(1 + cos2x) + (3/8)(1 - cos2x) + (√3/2) sin2x]
= (104) * [(1/2) - (1/4) cos2x + (√3/2) sin2x]
= (104) * [(1/2) + (1/4) (12/13√3) + (√3/2) * (11/13)]
= 52 + 30√3

=> area of the trapezium
= area of the square - area of equilateral triangle - areas of the two right triangles
= 52 + 30√3 - 22 - 23 - 26√3
= 7 + 4√3.

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Q.450. To find the equation of a non-standard parabola.

Question 450.
Determine the general equation for the parabola with its focus is located on (6,4) and that its directrix equation is 4x + y -6 = 0.

Answer 450.
Parabola is defined as a set of points equidistant from a given point, called focus and a given line, called directrix.
If P (x, y) is a point on the parabola,
its distance from the focus (6, 4) = √[(x - 6)^2 + (y - 4)^2]
and its distance from the directrix 4x + y - 6 = 0 is
l4x + y - 6l / √(4^2 + 1^2)
=> √[(x - 6)^2 + (y - 4)^2] = l4x + y - 6l / √(4^2 + 1^2)
=> 17 [(x - 6)^2 + (y - 4)^2] = (4x + y - 6)^2
=> 17 (x^2 + y^2 - 12x - 8y + 52) = (16x^2 + y^2 + 36 + 8xy - 12y - 48x)
=> x^2 + 16y^2 - 8xy - 156x - 124y + 848 = 0.

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Q.449. Area of a quadrilateral.

Question 449.
In quadrilateral ABCD it's known that:
AC * BD = x and AB² - BC² + CD² - DA² = y
Find area of quadrilateral in terms of x and y.

Answer 449.
Refer to the following figure.

Let A(0, 0) be at the origin.
Let X-axis be along AC and C = (m, 0)
Let B = (a, b) and D = (c, d)

AC * BD = x
=> m * BD = x
=> BD^2 = x^2/m^2
=> (a - c)^2 + (b - d)^2 = x^2/m^2 ... ( 1 )

AB^2 - BC^2 + CD^2 - DA^2 = y
=> a^2 + b^2 - (m - a)^2 - b^2 + (m - c)^2 + d^2 - c^2 - d^2 = y
=> 2m (a - c) = y

Plugging (a - c) = y/2m in ( 1 ),
y^2/4m^2 + (b - d)^2 = x^2/m^2
=> l b - d l = √[x^2/m^2 - y^2/4m^2]

Area of the quadrilateral
= (1/2) AC * l b - d l
= (1/2) m * √[x^2/m^2 - y^2/4m^2]
= (1/2) √[x^2 - y^2/4)].

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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.447. Elimination of θ from two equations.

Question 447.
Eliminate θ from (x/a) cos θ + (y/b) sin θ = 1 and xsin θ - y cos θ  = √(a² sin² θ + b² cos²θ).

Answer 447.
(x/a) cosθ + (y/b) sinθ = 1
=> (x/a)^2 cos^2 θ + 2(xy/ab) sinθ cosθ + (y/b)^2 sin^2 θ = 1
=> (x/a)^2 + 2(xy/ab) tanθ + (y/b)^2 tan^2 θ = sec^2 θ
=> (x/a)^2 + 2(xy/ab) tanθ + (y/b)^2 tan^2 θ = 1 + tan^2 θ
=> [(y/b)^2 - 1] tan^2 θ + 2(xy/ab) tanθ + [(x/a)^2 - 1] = 0
=> [(y^2 - b^2)/(x^2 - a^2)] * (a/b)^2 tan^2 θ + [2xy/(x^2 - a^2)] * (a/b) tanθ + 1 = 0 ... ( 1 )

xsinθ - ycosθ =√(a² sin² θ + b² cos² θ)
=> x^2 sin^2 θ - 2xysinθ cosθ + y^2 cos^2θ = a^2 sin^2 θ + b^2 cos^2 θ
=> x^2 tan^2 θ - 2xy tanθ + y^2 = a^2 tan^2 θ + b^2
=> (x^2 - a^2) tan^2 θ - 2xy tanθ + (y^2 - b^2) = 0
=> [(x^2 - a^2)/(y^2 - b^2)] tan^2 θ - [2xy / (y^2 - b^2)] tanθ + 1 = 0 ... ( 2 )

Comparing eqns. ( 1 ) and ( 2 ),

using coefficients of tanθ and constant terms,
[2xy/(x^2 - a^2)] * (a/b) / [- 2xy / (y^2 - b^2)] = 1
=> [1/(x^2 - a^2)] * (a/b) = - 1/(y^2 - b^2)
=> - a (y^2 - b^2) = b(x^2 - a^2)
=> bx^2 + ay^2 = ba^2 + ab^2
=> x^2/a + y^2/b = a + b.

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Q.446. Trigonometric Equation.

Question 446.
Solve 2sin(10º)sin(20º + θ) = sin(θ) for 0º < θ < 90º.

Answer 446.
2sin(10º)sin(20º + θ) = sin(θ)

=> cos(θ + 10º) - cos(θ + 30º) = sinθ
=> cos(θ + 10º) = cos(θ + 30º) + cos(90º - θ)
=> cos(θ + 10º) = 2cos60º cos(θ - 30º)
=> cos(θ + 10º) = cos(θ - 30º)
=> θ + 10º = ± (θ - 30º)
=> 2θ = 20º ... [taking the -ve sign on RHS as +ve sign gives no result]
=> θ = 10º.

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Q.441. Tangents - Co-ordinate geometry.

Question 441.
If  from  any point  on  the  common  chord  of  two  intersecting  circles,  tangents  are  drawn  to  the circles,  prove  that they are of  equal  lengths.

Answer 441.
Let x^2 + y^2 + 2g1x + 2f1y + c1 = 0
and x^2 + y^2 + 2g2x + 2f2y + c2 = 0 be any two circles.
Their common chord is obtained by taking the difference of the equations and is
2(g1 - g2)x + 2(f1 - f2)y + c1 - c2 = 0
Let (h, k) be any point external to the two circles on the above common chord.
=> 2(g1 - g2)h + 2(f1 - f2)k + c1 - c2 = 0 ... ( 1 )

If L1 and L2 are the lengths of tangents from (h, k) to the given circles,
L1^2 = h^2 + k^2 + 2g1h + 2f1k + c1 and
L2^2 = h^2 + k^2 + 2g2h + 2f2k + c2
=> L1^2 - L2^2
= 2(g1 - g2)h + 2(f1 - f2)k + c1 - c2
= 0 ... [from eqn. ( 1 )]
=> L1^2 = L2^2
=> L1 = L2.

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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.433. 3-D Trigonometric application.

Question 433.

The diagram shows a solid cylindrical post with a slanted elliptical face at an angle of θ to the diameter of the cylinder. Diameter DB = 10cm,

Height BC = 100cm and

0º < θ < 90º.

The red route (straight lines from A to B, then B to C) and

the blue route (straight line from A to C around half the circumference of the cylinder) are the same length. 

Prove that θ = π/2 - 2tan^-1 [(π^2 - 4)/80]
and find its value.

Answer 433.

Cutting open the cylinder at BC,

AC^2 = (5π)^2 + (100 + 10tanθ)^2

(AB+BC)^2 = (10secθ + 100)^2

AB+BC = AC

=> (AB+BC)^2 = AC^2

=> (100 + 10secθ)^2 = 25π^2 + (100 + 10tanθ)^2

=> 2000secθ + 100sec^2 θ = 25π^2 + 2000tanθ + 

      100tan^2 θ

=> 2000 (secθ - tanθ) = 25π^2 - 100

=> secθ - tanθ = (π^2 - 4)/80 ... ( 1 )

secθ - tanθ = (1 - sinθ)/cosθ
                      = [cos(θ/2) - sin(θ/2)]^2 / [cos^2 (θ/2) - sin^2 (θ/2)]

                      = [cos(θ/2) - sin(θ/2)] / [cos(θ/2) + sin(θ/2)]
                      = [1 - tan(θ/2)] / [1 + tan(θ/2)]
                      =  tan(π/4 - θ/2)   ... ( 2 )

From ( 1 ) and ( 2 ),

=> tan(π/4 - θ/2) = (π^2 - 4)/80

=> θ = π/2 - 2tan^-1 [(π^2 - 4)/80]
=> θ = 81.6°

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Q.431. Root of a Complex Number - De Moivre's Theorem

Question 431.
Z^10 = -1 (Z is a complex number).
Prove that the answers Z1, Z2, Z3....Z10 are Geometrical progression.

Answer 431.
Z^10 = -1
=> Z
= (-1)^(1/10)
= (cosπ + isinπ)^(1/10)
= [cos(2k+1)π + i sin(2k+1)π]^(1/10)
= cos (2k+1) (π/10) + i sin(2k+1) (π/10)
k = 10, 1, 2, 3, 4, .... 9 give the values of Z1, Z2, Z3, ... Z10
For them to be in G.P., we need to prove that the ratio of two successive roots is the same

Let Zn and Zn+1 be two successive roots
=> Zn = cos(2n-1) (π/10) + i sin(2n-1) (π/10)
and Zn+1 = cos(2n+1) (π/10) + i sin(2n+1) (π/10)
=> Zn+1 / Zn
= [cos(2n+1) (π/10) + i sin(2n+1) (π/10)] / [cos(2n-1) (π/10) + i sin(2n-1) (π/10)]
= cos(π/5) + isin(π/5)
As this is independent of n, the ratio is the same for any two successive terms
=> Z1, Z2, Z3, ... Z10 are in G.P.

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