Author Topic: Weird Integration/Trigonometry problem. (Read 2849 times) Tweet Share

lwine · « **on:** June 19, 2008, 08:07:06 pm »

Hi VN,

I came across this problem:

$\int\dfrac{2sin\sqrt{x}}{\sqrt{x}sec\sqrt{x}}dx$

I initially solved it:

$\int\dfrac{2sin\left(\sqrt{x}\right)cos\left(\sqrt{x}\right)}{\sqrt{x}}dx$
$\Rightarrow\int\dfrac{sin\left(2\sqrt{x}\right)}{\sqrt{x}}dx$
$\mbox{Let }u=2\sqrt{x}\Rightarrow\dfrac{du}{dx}=\dfrac{1}{\sqrt{x}}$
$\mbox{Therefore, }\int sinu\ du=-cosu+c\Rightarrow -cos\left(2\sqrt{x}\right)+c$

However, the suggested solution is $2\left(sin\sqrt{x}\right)^{2}$ , which I also managed to get by making the substitution $u=sin\sqrt{x}\Rightarrow\dfrac{du}{dx}=\dfrac{cos\sqrt{x}}{2\sqrt{x}}$
And then solving for $4\int u\ du = 2u^{2} = 2\left(sin\sqrt{x}\right)^{2}$

Using trig identities, I get $2\left(sin\sqrt{x}\right)^{2}=1-cos\left(2\sqrt{x}\right)}\not=-cos\left(2\sqrt{x}\right)$

I've been through this a few times; what is wrong with my initial approach?

Thanks!

/0 · « **Reply #1 on:** June 19, 2008, 08:10:48 pm »

The constant

$\int \frac{2\sin{\sqrt{x}}}{\sqrt{x}\sec{\sqrt{x}}}\,dx = -\cos{(2\sqrt{x})}+C = 2\sin^2{\sqrt{x}}+D$

$1-\cos(2\sqrt{x}) = 2\sin^2{\sqrt{x}}$

So let $C = D+1$

Mao · « **Reply #2 on:** June 19, 2008, 08:13:17 pm »

Quote from: lwine on June 19, 2008, 08:07:06 pm

Using trig identities, I get $2\left(sin\sqrt{x}\right)^{2}=1-cos\left(2\sqrt{x}\right)}\not=-cos\left(2\sqrt{x}\right)$

I've been through this a few times; what is wrong with my initial approach?

Thanks!

I made that oversight a while back

at the step of integration, you have arrived at $-cos\left(2\sqrt{x}\right)+c_1$

the final answer indicates $2(\sin\sqrt{x})^2+c_2$

if you equate these two terms, you'll just get $c_2+1=c_1$ , but since our cs are really just constants of integration [that can be any number], it doesnt actually matter at all.

it creeps up a bit in integration of trig functions, but these two are essentially the same. the constant is simply just "absorbed" into the constant of integration

lwine · « **Reply #3 on:** June 19, 2008, 08:15:15 pm »

Oh wow right! Yeah, of course!

Ahaha I feel kind of silly now (especially after typing all that latex).

Thanks Divideby0, and Mao (again).

Also, I guess its pretty useful to actually make the realisation that the constant is in fact arbitrary, from a problem like this, rather than encounter in an exam and (for some reason or other) get stuck.

//Edit: does this mean that it is safe to ignore any constants resulting from manipulating an indefinite integral by "absorbing" it into c?

lwine · « **Reply #4 on:** June 19, 2008, 11:08:48 pm »

Hi again,

I just attempted solving the following integral:

$\int\dfrac{1}{1+\sqrt{x}}dx$

I've tried multiplying the denominator by its conjugate, but no matter what I do, I always seem to get stuck. The solution is $2\sqrt{x}-2log_e\left(1+\sqrt{x}\right)$

Any hints/solutions would be greatly appreciated.

Thanks!

Neobeo · « **Reply #5 on:** June 19, 2008, 11:46:48 pm »

Quote from: lwine on June 19, 2008, 11:08:48 pm

Hi again,

I just attempted solving the following integral:

$\int\dfrac{1}{1+\sqrt{x}}dx$

I've tried multiplying the denominator by its conjugate, but no matter what I do, I always seem to get stuck. The solution is $2\sqrt{x}-2log_e\left(1+\sqrt{x}\right)$

Any hints/solutions would be greatly appreciated.

Thanks!

Try the substitution $u=1+\sqrt{x}$

Then we have $2(u-1)\mbox{du}=\mbox{d}x$

Giving us $\int\frac{\mbox{d}x}{1+\sqrt{x}} = \int\dfrac{2u-2}{u}\mbox{d}u$

lwine · « **Reply #6 on:** June 20, 2008, 06:11:29 pm »

Ah thanks Neobeo, I didn't see that.

Another problem I'm stuck on if anybody is feeling helpful (I've got a sheet full of difficult problems that I've been grinding through for the past two days in preparation for a SAC):

$\int_{0}^{1}\sqrt{\dfrac{1-x}{1+x}}dx$
The solution is $\dfrac{\pi-2}{2}$

Thanks so much!

Mao · « **Reply #7 on:** June 20, 2008, 06:16:12 pm »

$\int \sqrt{\frac{1-x}{1+x}}\; dx=\int \sqrt{\frac{(1-x)^2}{1-x^2}}\; dx=\int \frac{1}{\sqrt{1-x^2}}\; dx\; -\; \int \frac{x}{\sqrt{1-x^2}}\;dx$

first part is arcsin, second part can be computed with the substitution method

lwine · « **Reply #8 on:** June 20, 2008, 06:26:13 pm »

What the hell!

I spent half an hour trying to figure that one out. I can't believe I didn't see that. Oh wow.

Thanks Mao!

lwine · « **Reply #9 on:** June 22, 2008, 10:59:52 pm »

Hi guys,

I almost cried when I saw this one.

Any hints/solutions will be a great help!

kj_ · « **Reply #10 on:** June 23, 2008, 06:43:05 am »

Quote from: lwine on June 22, 2008, 10:59:52 pm

Hi guys,

I almost cried when I saw this one.

(Image removed from quote.)

Any hints/solutions will be a great help!

hints? (seeing as you have the answer already), find $\frac{dt}{dx}$ , change the terminals so your integrand is in terms of t, sub $\frac{1}{t} + 1$ in wherever you see $x$

Mao · « **Reply #11 on:** June 23, 2008, 06:34:02 pm »

Quote from: lwine on June 22, 2008, 10:59:52 pm

Hi guys,

I almost cried when I saw this one.

(Image removed from quote.)

Any hints/solutions will be a great help!

let $x+1=\frac{1}{t}$ , when x=0, t=1, x=1, t=0.5

$\frac{dx}{dt}=\frac{d}{dt}\left(\frac{1}{t}-1\right)=-\frac{1}{t^2}$

$\int_0^1 \frac{1}{(x+1)\sqrt{8x^2+14x+5}}\; dx = \int_0^1 \frac{1}{(x+1)\sqrt{8x(x+1)+5(x+1)+x}}\; dx$

$\implies \int_1^{0.5} -\frac{1}{t^2}\cdot \frac{1}{t^{-1}\cdot \sqrt{8(t^{-1}-1)t^{-1} + 5t^{-1} + (t^{-1}-1)}}\; dt$

$\implies \int_{0.5}^1 \frac{1}{\sqrt{t^2 \cdot (8t^{-2}-8t^{-1} + 5t^{-1} -1 + t^{-1})}}\; dt$

$\implies \int_{0.5}^1 \frac{1}{\sqrt{8-2t -t^2}}\; dt$

$\implies \int_{0.5}^1 \frac{1}{\sqrt{9-(t+1)^2}}\; dt$

$\implies \left[ \sin^{-1}\left(\frac{t+1}{3}\right)\right]_{0.5}^1$

$\implies \sin^{-1}\left(\frac{2}{3}\right) - \sin^{-1}\left(\frac{1.5}{3}\right)$

$\implies \sin^{-1}\left(\frac{2}{3}\right) - \frac{\pi}{6}$

* Mao *at last!*

lwine · « **Reply #12 on:** June 23, 2008, 11:36:07 pm »

Oh, mao, you're amazing.

Thanks alot.

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Author Topic: Weird Integration/Trigonometry problem. (Read 2849 times) Tweet Share

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Weird Integration/Trigonometry problem.

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