Humphrey77: solve for h in the equation : 6(raise to power h) -31(2 raise to power h ) +32 is equal to zero.

Humphrey77: SOLVE FOR h in the equation : 6(raise to power h) -31(2)raise to power h+32 is equal to zero ( HINT : USE LINEAR APPROX)
:* HAPPY!

benbuks: xmas. question. integrate sec^3 x dx

olasesi: find the SUM of the nth term of the sequence.........4+44+444+4444+44444+444444+......

it was given as an assignment for us back then in pre-degree........
.......only me got the answer to it .......i felt fulfilled then.........
........I HOPE THERE ARE OTHER PERSONS WHO CAN SOLVE IT.......

doubleDx:

∫sec³x dx
∫sec³x dx can be rewritten as =>

∫sec²x sec x dx, using integration by parts=>

Let's put

u = sec x, differentiating yields du = secxtanx dx.

Again let's put dv = sec² x dx, integrating yields => v = tan x. Now the formula for integration by parts is =>

∫udv = uv - ∫vdu, substituting yields=>

∫sec³x dx = sec x tan x - ∫tan x (sec x tan x)dx
∫sec³x dx = sec x tan x - ∫tan² x sec x dx

Remember from trig identities that => tan² x = sec² x - 1, substituting yields=>

∫sec³x dx = sec x tan x - ∫sec x (sec² x - 1) dx
∫sec³x dx = sec x tan x - ∫sec³ x dx + ∫sec x dx

Collecting like terms by taking the bolded part to the LHS yields =>

∫sec³x dx + ∫sec³ x dx = sec x tan x + ∫sec x dx
2∫sec³x dx = sec x tan x + ∫sec x dx
2∫sec³x dx = sec x tan x + ln|sec x + tan x| + C
:. ∫sec³x dx = (1/2){sec x tan x + ln|sec x + tan x|} + C

Merry Christmas and Happy New Year in advance to Y'all!! 1luv

rashywire: pls help luk @ dis question?
d sum of d first ten terms of an a.p is -55 while d sum of d next 7term is -217. find (i) d sum of d first 17th terms (ii) first term n d common difference

benbuks: great work master..c as u kill d tin...na cheep 1 sha..

rashywire: pls help luk @ dis question?
d sum of d first ten terms of an a.p is -55 while d sum of d next 7term is -217. find (i) d sum of d first 17th terms (ii) first term n d common difference

doubleDx:

Solution to Question (ii)

Sum of n terms is given by =>

S_n = n/2{2a + (n - 1)d}
Sum of the first ten terms (S₁₀) => -55, n = 10

:.
10/2{2a + (10 - 1)d} = -55
5(2a + 9d) = -55
2a + 9d = -11.....equation (1)

Now, the sum of the next 7th terms (S₇) => -217. Which means the first term (A) in terms of the general 1st term (a) and common difference (d) = 11th term of the general series, which is => (a + 10d) and n = 7

:. S_n = n/2{2A + (n - 1)d}

7/2{2(a + 10d) + (7-1)d} = -217

7{2a + 20d + 6d) = -217(2)

2a + 26d = -31(2)
a + 13d = -31

a = -31 - 13d ......equation (2)

substitute a in equation (1)

2a + 9d = -11
2(-31 - 13d) + 9d = -11
-62 - 26d + 9d = -11
-17d = 62 - 11
d = -51/17
d = -3

Substitute for d in equation (2) to get a.

a = -31 - 13d
a = -31 - 13(-3)
a = -31 + 39
a = 8

:. a = 8 and d = -3
-----------------------

Question (i)

S_n = n/2{2a + (n - 1)d}

Now, a = 8, d = -3, n = 17

sum of first 17 terms (S₁₇) = 17/2 {2 ( 8 ) + (17 - 1)(-3)}
S₁₇ = 17/2 (16 - 48)
S₁₇ = (17/2)(-32)
S₁₇ = 17(-16)
S₁₇ = -272

2nioshine: Have gat two pending questions on laplace transform. Any attempt to it will be so mush appreciated..lets warm up with this

1. Show that:
Sin(45-x)/cos2x=sec(45-x)/2

2. If dx=-_/(100-y²)dy/y
show that x=-_/(100-y²) + 10 ln[10+_/(100-y²)/y] +c

pls the solution to Q2 is very important and really needed...i will start giving my full contributions in no time.Any atempt to the above is welcomed...

doubleDx:

Solution to Q2

dx = - √(100 - y²)dy/y

To compute x from the above, we have to integrate both sides =>

∫dx = ∫{-√(100 - y²)dy/y}
x = - ∫{√(100 - y²)dy/y}

Put =>

dv = dy/y integrating yields => v = lny

Put u = - √(100 - y²), differentiating yields => du = ydy/√(100 - y²)

Using integration by parts =>

∫udv = uv - ∫vdu

x = lny(- √(100 - y²)) - ∫ylnydy/√(100 - y²)
x = -lny√(100 - y²) - ∫ylnydy/√(100 - y²)..........(1)

Now for the integrand ∫ylnydy/√(100 - y²)

Let's put u = lny, differentiating yields => du = dy/y

dv = ydy/√(100 - y²), integrating yields => v = √(100 - y²)

∫udv = uv - ∫vdu

∫ylnydy/√(100 - y²) = lny√(100 - y²) - ∫√(100 - y²)dy/y

substituting the above in eqn (1) yields =>

x = -lny√(100 - y²) + lny√(100 - y²) + ∫√(100 - y²)dy/y}
x = -lny√(100 - y²) + lny√(100 - y²) + ∫√(100 - y²)dy/y
x = ∫√(100 - y²)dy/y.................(2)

To compute => ∫√(100 - y²)dy/y

Put y = 10sin(u), so that u = sin^-1(y/10); Let => dy = 10cos(u)du

Substituting yields =>

:. ∫√(100 - y²)dy/y =>

=> ∫√{100 - 10²sin²(u)}{10cos(u)du}/10sin(u)
=> ∫√(100){1 - sin^(u)}{10cos(u)du}/10sin(u)
=> 10∫√cos²(u).{10cos(u)du}/10sin(u)
=> 10∫cos(u).cot(u)
=> 10∫cos(u)cot(u)

Now, cos(u)cot(u) => cos²(u)/sin(u)
=> {1 - sin²(u)}/sin(u)
=> 1/sin(u) - sin²(u)/sin(u)
=> cosec(u) - sin(u)

:. ∫√(100 - y²)dy/y
=> 10∫cosec(u) - 10∫sin(u)
=> - 10ln{cot(u) + cosec(u)} + 10cos(u)
substituting for u = sin^-1(y/10) and evaluating yields =>

∫√(100 - y²)dy/y = √(100 - y²) - 10ln{[10 + √(100 - y²)]/y} + C

Since =>
x = ∫√(100 - y²)dy/y
x = √(100 - y²) - 10ln{[10 + √(100 - y²)]/y}
x = - 10ln{[10 + √(100 - y²)]/y} + √(100 - y²) + C . Proven QED.

2nioshine:
The lost of the negative sign from the actual question in arriving at the first bolded actually afected d final solution. compare eqn 2 to the actual quest. but that not witstanding ur soln was xo on point

The second bolded part was indeed very helpful...

Unique approach,nice presentation.tnks boss

doubleDx:

Thanks for the observation, I typed the solution in a rush...I will cross check it later. I think I made a little mistake in the expansion before arriving at equation (2). 1luv

Edited: I have figured it out. The integration by parts wasn't necessary....I should have used substitution directly! Yes, if you apply substitution method carefully you will arrive at the correct answer. I will fix my solution later.

Here it goes again =>

Solution to Q2

dx = - √(100 - y²)dy/y

To compute x from the above, we have to integrate both sides =>

∫dx = ∫{-√(100 - y²)dy/y}
x = - ∫{√(100 - y²)dy/y}

To compute the RHS,

Put => y = 10sin(u) and sin(u) = y/10, so that u = sin^-1(y/10); Let => dy = 10cos(u)du

:. -∫√(100 - y²)dy/y =>

=> -{∫√{100 - 10²sin²(u)}{10cos(u)du}/10sin(u)}
=> -∫√(100){1 - sin^(u)}{10cos(u)du}/10sin(u)
=> -10∫√cos²(u).{10cos(u)du}/10sin(u)
=> -10∫cos(u).cot(u)du
=> -10∫cos(u)cot(u)du

Now, cos(u)cot(u) => cos²(u)/sin(u)
=> {1 - sin²(u)}/sin(u)
=> 1/sin(u) - sin²(u)/sin(u)
=> cosec(u) - sin(u)

:. -∫√(100 - y²)dy/y => -10∫cos(u)cot(u)du
=> -10∫cosec(u)du + 10∫sin(u)du}
=> - 10cos(u) + 10ln{cot(u) + cosec(u)}
Substituting for u = sin^-1(y/10) and evaluating yields =>

Since => sin(u) = y/10
Then => cos(u) = √(1 - sin²(u))
=> cos(u) = √(100 - y²)/10

=>Cosec(u) = 1/sin(u)
=>Cosec(u) = 10/y

=>Cot(u) = cos(u)/sin(u)
=>Cot(u) = (√(100 - y²)/10).(10/y)
=>Cot(u) = √(100 - y²)/y


:.=> - 10cos(u) + 10ln{cot(u) + cosec(u)}

=> -10√(100 - y²)/10 + 10ln{√(100 - y²)/y + 10/y}
=> -√(100 - y²) + 10ln{[10 + √(100 - y²)]/y} + C

:. x = - ∫{√(100 - y²)dy/y} => -√(100 - y²) + 10ln{[10 + √(100 - y²)]/y} + C

Pat £inst£in:

The coefficient of x^4 in the expansion of (1 - x + x^2)^4 is __