Integration Challenge?

Question

I have 5 integrals that I'm working on and I need everyone's help on these problems thank you very much who can answer these.
#1. The indefinite integral of (dx)/(1+sinx+cosx). (Hint: Let u= tan(x/2).
#2. The indefinite integral of (sqrt Tanx) dx
#3. The indefinite integral of sqrt(x-1)/sqrt(x+1) * 1/x^2 dx
#4. The indefinite integral of (dx)/(x^6-1).
#5. The indefinite integral of (x^2-1)/(x^4+3x^3+5x^2+3x+1) dx

Please I want to know every little detail of what you did in each part of the problem not just the answer. Once again I thank you for helping me if anyone should solve these integrals.

Answer 1

Wow! Any one of these integrals would be a great
final problem for an integration bee!
I have worked out nos 1,2,4 and 5.
I don't have time to type out all the steps right now,
but let me give you hints for these 4 and I'll come back
later and fill in all the details.
1. You do this by half-angle substitution.
As you wrote, let u = tan(x/2)
Then x = 2 arctan u
dx = 2/(1+ u²)
Now you need to find cos x and sin x.
Use the double angle formula for cos x:
cos x = 2 cos² x/2 - 1
= 2/ sec² x/2 - 1
= 2/(1+ tan² x/2) - 1
= 2/ (1 + u²) - (1+u²)/(1+u²) = (1-u²)/(1+u²)
To find sin x make a right triangle with hypotenuse
1 + u² and leg 1-u².
By Pythagorean theorem the third side is 2u
and sin x = 2u/(1+u²).
Now plug these into your integral and grind!

2. ∫ √tan x dx
Let u = tan x, x = arctan u
dx = du/(1+u²)
So we have to compute
∫ √u/(1+u²) du
Let u = t², du = 2tdt
So our integral finally becomes
∫ 2t² dt /(t^4+1). (*)
What to do witth (*)?
The secret here is that t^4 + 1 = (t²+√2t+1)(t²-√2t+1)
So plug this into (*) and use partial fractions.
You should get 4 terms, 2 of which involve logarithms
and the other 2 involve arctangents.

4. Factor x^6 -1 as (x³-1)(x³+1) = (x-1)(x+1)(x²+x+1)(x²-x+1)
Separate the denominator into partial fractions and
proceed.

5. Not easy to spot how to do this one!
The quartic is irreducible over the rationals,
so it's hopeless to try to factor it.
The saving grace is that the denominator
is a reciprocal quartic. In other words
the coefficients are the same in either direction.
It also means that if r is a zero of this polynomial
so is 1/r.
A trick that often works with these type of
polynomials is to use the substitution u = x+1/x.
So let's divide everything by x² and look at
∫ (1 - 1/x²) dx / (x² + 3x + 5 + 3/x + 1/x²)
and the denominator can be written as
(x+1/x)² + 3(x+1/x) + 3.
So let u = x+1/x. Then du = (1 - 1/x²) dx.
So now we have to compute
∫ du /(u²+3u+3)
Next, rewrite this as
∫ du/ (u²+ 3u + 9/4) + 3/4,
i.e complete the square in the denominator.
This is
∫ du/[ (u+ 3/2)² + (√3/2)²)
which works out to
2/√3 * arctan( (2u+3)/√3)
Now back substitute to get your final answer.
Whew!!

Answer 2

The only one answer is the best one even if it doesn't answers the question. Right?