help please on integral calculus?

Question

i seem to forgot what formula is to use.... can you please tell me what will i do with the (integration of the square root of tanx dx)?

Answer 1

The last person's reasoning appears correct but I think that their final line contains something that is not readily integrable even with partial fractions. I must admit that this one has me stumped. You've ruined my evening, as I think that I should be able to do it, so will keep trying. If you made it up rather than it being a book question (or copied it wrongly) it is possible that it is a function you just can't integrate like e^(x^2).

Answer 2

Hint: Let u = tan x, x = arctan u, dx = du/(u²+1).
Then you have to integrate sqrt(u)/(u²+1).
Now try another substution: t = sqrt(u), u = t^2, du = 2t dt.
So our final problem is to integrate
2t²/(t^4+1). Now we must use partial fractions to complete the work.
Here is how to start using partial fractions:
You have to write the denominator as t^4 + 1= t^4 + 2t² + 1 - 2t².
Now factor this as (t² +√2 t + 1)(t² -√2 t + 1).
Next, let's get the partial fraction decomposition.
We have
2t² = (At+B)(t² - √2 t +1) + (Ct+D)(t² + √2 t + 1).
Now equate coefficients of like powers of t:
Constant terms: B+D=0.
t³ terms: A+C=0
t terms: A -B√2 + C + D√2 = 0,
which gives B = D = 0.
Finally, using these last results, and equating the
coefficients of the t² terms, we get
A(-2√2) = 2
A = -1/√2, C = 1/√2
Rewriting this, we see that our final task is to integrate
1/√2( -t / (t² - √2 t + 1) + t / (t² + √2 t + 1) ).
Now we want to rewrite the numerator of each fraction
to get the derivative of the denominator plus a constant
on top. So we will rewrite the integrand as
-1/2√2( (2t - √2 + √2)/ (t² - √2 t + 1)
+ 1/2√2(2t + √2 - √2)/ (t² + √2t + 1) )
Integrating the first two terms of each fraction,
we get
1/(2√2)(-log (t² - √2 t + 1) + log(t² + √2 t + 1) ).
We still have to integrate
1/2√2( -√2 /( t² - √2t + 1) + √2 / (t² + √2t + 1). (*)
Let's do the first integral. The second one is done
exactly the same way.
We have to compute
-1/2 ∫ 1/ (t² -√2 t + 1) dt.
Let's complete the square in the denominator and write it as
-1/2 ∫ 1/ (t² -√2 t + 1/2 + 1/2) dt =
-1/2 ∫ 1/[ (t- √2/2 t)² + 1/2] dt.
Now let s = t - √2/2, ds = dt.
Then we have
-1/2 ∫ ds/ (s² + 1/2 ) (**)
From elementary calculus, we recall that
∫ ds / (s² + a²) = 1/a arctan(s/a)
So, applying this formula to (**) with a = 1/ √2, we
see that (**) equals
-√2/ 2 arctan(√ 2 s) =
-1/√2 arctan (√2 t -1).
In the same way, the second integral of (*) equals
1/√2 arctan(√2 t + 1).
Putting it all together, we finally have
∫ 2t² dt/ (t^4 + 1) =
1/(2√2)[ log (t² - √2 t + 1) - log(t² + √2 t + 1 )
- 2 arctan (√2 t -1) + 2 arctan(√2 + 1)] + C.
Now you can back substitute to get the answer
for the original problem.
Whew, what a complicated integral!!
Hope this was helpful to you!