₦airaland Forum

Mikebis: Cn any1 pls help me wt dis questn.1.fnd dy/dx of (x^2+1)^x^2..(2)solv simultaneously x+y+z=8..eqn1
x^2+y^2+z^2=72..eqn2
x^3+y^3+z^3=216..eqn3.(3)prv dat x^3+y^3+z^3=(x+y+z)^2(x^2+y^2+z^2-xy-yz-zx)..(4)when x=3^2/3+3^-2/3 prv dat 9x^3-27x=82..pls help me,nd as do it may GOD increase ur knowledge.AMEN

mathematician: for the first question, we seperate the two parts of the qustion into different functions i.e
y= 2^x...............(1)
y=2x+9..................(2)
We can only solve this graphical, so by plottig the graph, we will see that x is approximately 4.13
pls manage the primitiv egraph i attached...its a rough one

mathematician: doubledx u are totally wrong, now my working the eqn. 1/(cosx+1)² can also be written as (cosx+1)^-2, this means we have to use integration by algebraic subtitution. let u=(cosx+1), then du/dx= (-sinx), then then §(cosx+1)^-2.dx, becomes 1/(-sinx)§u^-2.du, after the integration we have /(-sinx)*(-u^-1) by replacing 'u' for its original value and multiplying, we have (sinx.cosx+sinx)^-1 + c or 1/(sinx.cosx+sinx)+c i hope u can see your mistake

doubleDx: masperano wasn't cricticizing, I made a little typo-error and he made an honest observation, nothing more....

doubleDx: masperano wasn't cricticizing, I made a little typo-error and he made an honest observation, nothing more....

masperano: double dx how can -4cos^2(x/2) + 4cos^2x= 0 ( please check ur workings again)

doubleDx: A little error in my previous post, I have checked all steps on my PC & my workings are unarguably right, so I'm re-posting.... Check below carefully and re-quote me! You can post your own method for the understanding of other viewers too; but mind is correct too and cannot be disproved....



The 1st thing to do here is to simplify the numerator = (cos x + 1)²
Expanding yields=>
Cos²x + 2cosx + 1 .....(1)

Now, from trig identities, we know that cos(A + B ) = cosAcosB - sinAsinB

So that cosx = cos (x/2 + x/2)
= cos(x/2).cos(x/2) - sin(x/2).sin(x/2)
= cos²(x/2) - sin²(x/2) .... (2)

But sin²ø = 1 - cos²ø
:. Replacing ø = x/2 and substituting back in (2) yields =>

cos x = cos²(x/2) - [1 - cos²(x/2)]
cos x = 2cos²(x/2) - 1 ... Half angle formula!

Having gotten the xpression for cos x, substituting back in eqn (1) yields =>

[2cos²(x/2) - 1 ]² + 2[ 2cos²(x/2) - 1 ] + 1

Expanding =>
4cos⁴(x/2) - 4cos²(x/2) + 1 + 4cos²x - 2 + 1

Collecting like terms and simplifying yields =>
4cos^4(x/2) - 4cos^2(x/2) + 4cos^2x - 2 + 2
= 4cos^4(x/2)
Hence the numerator =>
(cos x + 1)² = 4cos⁴(x/2)

So that the question ∫ [1/(1 + cos x)²] dx is reduced to = ∫ 1/4cos⁴(x/2)dx
= ∫ 1/4sec⁴(x/2) dx

∫1/4sec⁴(x/2) dx
1/4∫ sec² (x/2) sec²(x/2) dx

Put (x/2) = t

So that =>
dx = 2dt and =>
= 1/4∫ sec²(x/2) sec²(x/2) dx
= 1/4∫ sec²t sec²t .2dt

Remember that sec²t => tan²t + 1
Substituting this for one of the sec²t yields =>

1/2∫ (tan²t + 1) sec²t dt
1/2(∫ [sec²t tan²t + sec²t] ) dt
Now, spliting into two integrals yields =>
1/2[∫ sec²ttan²t dt + ∫ sec²tdt ]

Integrating separately =>
For=> ∫ sec²t tan²t dt

Put u = tan t
So that => du = sec²t dt
Substituting yields =>
∫ u² du
= (1/3) u³ + C
= (1/3) tan³t + C
For => ∫ sec²t dt
= tan t + C

Putting the two integrals together yields=>

= 1/2[(1/3) tan³t + tant]
= 1/6tan³t + 1/2tant + C
Remember that t = x/2
Putting it back into your answer yields =>
= 1/6tan³(x/2) + 1/2tan(x/2)+ C

∫[1/( cos x + 1)²]dx = 1/6tan³(x/2) + 1/2tan(x/2) + C

===============
You can check my answer by differentiating 1/6tan³(x/2) + 1/2tan(x/2) + C and you will arrive @ =>

1/( cos x + 1)²] or 1/4sec⁴(x/2)

doubleDx: The 1st thing to do here is to simplify the numerator = (cos x + 1)^2
Expanding yields=>
Cos^2x + 2cosx + 1 .....(1)

Now, from trig identities, we know that cos(A + B ) = cosAcosB - sinAsinB

So that cosx = cos (x/2 + x/2)
= cos(x/2).cos(x/2) - sin(x/2).sin(x/2)
= cos^2(x/2) - sin^2(x/2) .... (2)

But sin^2ø = 1 - cos^2ø
:. Replacing ø = x/2 and substituting back in (2) yields =>

cos x = cos^2(x/2) - [1 - cos^2(x/2)]
cos x = 2cos^2(x/2) - 1 ... Half angle formula!

Having gotten the xpression for cos x, substituting back in eqn (1) yields =>

[2cos^2(x/2) - 1 ]^2 + 2[ 2cos^2(x/2) - 1 ] + 1

Expanding =>
4cos^4(x/2) - 4cos^2(x/2) + 1 + 4cos^2x - 2 + 1

Collecting like terms and simplifying yields =>
4cos^4(x/2) - 4cos^2(x/2) + 4cos^2x - 2 + 2
= 4cos^4(x/2)
Hence the numerator =>
(cos x + 1)^2 = 4cos^4(x/2)

So that the question ∫ [1/(1 + cos x)^2] dx is reduced to = ∫ 1/4cos^4(x/2)dx
= ∫ 1/4sec^4(x/2) dx

∫1/4sec^4 x dx
1/4∫ sec^2 (x/2) sec^2 (x/2) dx

Put (x/2) = t

So that =>
= 1/4∫ sec^2 (x/2) sec^2 (x/2) dx
= 1/4∫ sec^2 t sec^2 t dt

Remember that sec^2t => tan^2 t + 1
Substituting this for one of the sec^2t.

1/4∫ (tan^2t + 1) sec^2t dt
1/4(∫ [sec^2t tan^2t + sec^2t] ) dt
Now, spliting into two integrals yields =>
1/4[∫ sec^2ttan^2t dt + ∫ sec^2tdt ]

Integrating separately =>
For=> ∫ sec^2t tan^2t dt

Put u = tan t
So that => du = sec^2t dt
Substituting yields =>
∫ u^2 du
= (1/3) u^3 + C
= (1/3) tan^3t + C
For => ∫ sec^2t dt
= tan t + C

Putting the two integrals together yields=>

= 1/4[(1/3) tan^3t + tant]
= 1/12tan^3t + 1/4tant + C
Remember that t = x/2
Putting it back into your answer yields =>
= 1/12tan^3(x/2) + 1/4tan(x/2)+ C

∫[1/( cos x + 1)^2]dx = 1/12tan^3(x/2) + 1/4tan(x/2) + C

echibuzor: Have you tried Quotient rule? The denominator is Function of a function...

phshawdy: nope its not, it'll be on permanent record/eligibility slip...

Zico0: Please who can integrate these derivatives:
integrate tanxdx

integrate 1/_/(x - 2)^2 + 25dx
i hope it is comprehensible.