Trigonometric substitution?

Question

Please help with this one too...

Evaluate using a trigonometric substitution:

Int(dx/(sqrt(2x^3 + 2x + 5)))

..thanks

Answer 1

∫ (1/ √(2x² + 2x + 5) dx )

To solve this, first complete the square in the denominator. Note that
2x² + 2x + 5 = 2(x² + x) + 5
= 2(x² + x + 1/4) + 5 - 1/2
= 2(x + 1/2)² + 9/2

∫ 1/√(2(x + [1/2])² + 9/2) dx

First, we use u-substitution.

Let u = x + [1/2]. Then
du = dx

∫ 1/√(2u² + (9/2)) du

Here is where we use trigonometric substitution. Your goal is to choose us such that inside of the radical equal to atan²(t) + a, so that you may exploit the identity tan²(t) + 1 = sec²(t).

Let u = [3/2] tan(t). Then
du = [3/2] sec²(t) dt

∫ 1/√( 2(9/4)tan²(t) + (9/2) ) [3/2] sec²(t) dt

∫ 1/√( (9/2)tan²(t) + (9/2) ) [3/2] sec²(t) dt

I'm going to fast-forward the details of this, and let you know that this eventually reduces to

(3/2)(√(2)/3) ∫ ( 1/sqrt(sec²(t)) sec²(t) dt )

(√(2)/2) ∫ (sec²(t)/sec(t) dt )

√(2)/2 ∫ ( sec(t) dt )

It is highly recommended that you memorize the integral for sec(t) dt, because it is one of the more nonintuitive integrals to solve for. The integral of sec(t) is ln|sec(t) + tan(t)|. Therefore our answer is

( √(2)/2 ) ln|sec(t) + tan(t)| + C

To convert this back in terms of u, remember that
u = (3/2)tan(t), so it follows that

tan(t) = 2u/3

Here's where we use trigonometry and SOHCAHTOA.
Think of a right angle triangle with t as the angle. Then, if

tan(t) = 2u/3 = opposite/adjacent.
opposite = 2u
adjacent = 3, so by Pythagoras,
hypotenuse = √( [2u]² + 3² ) = √(4u² + 9).

Since cos(t) = adjacent/hypotenuse,
sec(t) = hypotenuse/adjacent, so with tan(t) = 2u/3, and
sec(t) = √(4u² + 9)/3

( √(2)/2 ) ln|sec(t) + tan(t)| + C becomes

( √(2)/2 ) ln|√(4u² + 9)/3 + (2u)/3| + C

But, u = x + (1/2), so our final answer is

( √(2)/2 ) ln|√(4[x + (1/2)]² + 9)/3 + (2[x + (1/2)])/3| + C

Answer 2

Could it be that question should read:-
I = ∫1 / (2x² + 2x + 5)^(1/2).dx ?
If so:-
(2x² + 2x + 5)^(1/2)
= [2 (x² + x + 5/2)]^(1/2)
= [2 (x² + x + 1/4 - 1/4 + 5/2)]^(1/2)
= 2^(1/2).[(x + 1/2)² + 9/4]^(1/2)
I = (√2). ∫ 1 / [(x + 1/2)² + 9/4 ]^(1/2).dx
I = (√2).sinh^(-1)[(x + 1/2) / 9/4] + c
I = (√2).sinh^(-1) [(4x + 2) / 9 ] + c

Answer 3

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