Stupid calculus integration problem...?

Question

hello.....

wuz wonderin' if anyone can solve square root (( x^2 + 4 )) using trig identity?????????????????

im pullin my hairs out!!!

yah.. the indefinite integration of:

Square Root of (x^2 + 4)

using a trigonometric substitution

Answer 1

Ok, it's really roundabout, but I think I got it. Let's start with integration by parts, where u = sqrt(x^2+4) and dv = dx.

That tells us that:

Int[sqrt(x^2+4)] =
x*sqrt(x^2+4) - Int[(x^2)/sqrt(x^2+4)] =
x*sqrt(x^2+4) - Int[(x^2 + 4)/sqrt(x^2+4)] + Int[4/sqrt(x^2+4)]

So,
2Int[sqrt(x^2+4)] = x*sqrt(x^2+4) + Int[4/sqrt(x^2+4)]

So our problem has been reduced to Int[4/sqrt(x^2+4)], which we can now use trig sub on. Let x = 2tanu, dx = 2(secu)^2, and then we get:

Int[4/sqrt(x^2+4)] = Int[4secu] which we can look up or figure out to be 4ln|secu + tanu| =
4ln|sqrt((x^2)/4 + 1 + x/2|

So you original integral is:

(1/2)x*sqrt(x^2+4) - 2ln|sqrt((x^2)/4 + 1) + x/2|

This looks way messier than it should be, so you will want to work through the steps on your own to make sure I did not do something completely stupid.

Answer 2

I want to help you out without giving you the answer --

Remember the general form of trig substitution when a √ is involved:

√( b^2x^2 - a^2) => x = a/b sec (theta)

in your problem: ∫√x^2+4 dx, a =2 and b= 1

so try subbing x for 2sec(theta) and see what happens

Answer 3

this problem is what you called "inverse hyperbolic function"

can you remember the integrl of dx / sqr root of x^2 - a^2? ...well can remember now.

lets do it in that form..

convert the variables sqrt rt x^2 + 4 into exponential form.

it looks like this: intgrl of ( x^2 + 4 ) ^1/2 dx

now do it in fraction.

intgrl of dx / ( x^2 + 4 ) ^-1/2 ( notice the minus sign?)

bring it back to sqrt rt form.

intgrl of dx / -sqrt rt of ( x^2 + 2^2 ) : hence 2^2=4

remove the minus sign, but change the operation to minus.

now the equation is: intgrl of dx / sqrt rt of ( x^2 - 2^2 ) which is equal to.

cosh ^-1x/2 + C or ln | x + sqrt rt of (x^2 - 2^2) / 2 | + C answer. hope it helps

Answer 4

Let x = 2 sec Θ.

Then x² + 4 = 4sec² Θ + 4 = 4 (sec² Θ + 1) = 4 tan² Θ.

That means √(4 tan² Θ) = 2 tan Θ = 2 [sin Θ/cos Θ]

2 [sin Θ/cos Θ] = 2 (1/cos Θ)(sin Θ) Does this look familiar?

Now we let u = cos Θ and du = -sin Θ dΘ.

That means that (1/cos Θ)(sin Θ) dΘ is in the form (1/u) (-du), which is integrable.

So 2 (1/cos Θ)(sin Θ) dΘ is equivalent to -2 (1/cos Θ)(-sin Θ) dΘ.

Now we can integrate the last expression.

Int -2 (1/cos Θ)(-sin Θ) dΘ = -2 Int (1/cos Θ)(-sin Θ) dΘ =

-2 ln |cos Θ| + c, which is your final answer.

Answer 5

x^6

Answer 6

well I'm not sure what you are asking for when you say "trig identity"...x would equal 2i but but is there a sine or cosine or something more to the problem?...you can try and e-mail me I'm pretty good at math...

Answer 7

hm...i can solve it without a trigonometrical identity....

sqrt(x²+4)dx=
(x²+4)^1/2 dx

(x²+4)^1/2+1
=------------------ + C=
1/2+1

(x²+4)^3/2
= --------------- + C=
3/2

sqrt((x²+4)³)
=-------------------- + C
3/2

hope that's right ;)

Answer 8

what are we solving for? are you supposed to integrate?