# Thread: Evaluate the Indefinite Integral

1. ## Evaluate the Indefinite Integral

Hi,
I am having trouble evaluating the indefinite integral of the following problem using U-substitution. The answer at the back of the book is 2sin square root of t + C. I'm not sure what U should be. Any assistance will be appreciated. Thanks in advance.

1. The problem statement, all variables and given/known data

Evaluate the indefinite integral

2. Relevant equations

integral of cos * (square root of t) / (square root of t) dt

3. The attempt at a solution

integral of cos * (square root of t) / (square root of t) dt

Let U = ?

Lets say U = cos * (square root of t)

du = sin * (square root of t) dx

This does not make any sense to me. Im I going right? Where did I go wrong?

2. Edit: Nevermind, that's derivatives. Hold on.

3. $\displaystyle \cos(\sqrt{t}) = 1 - \frac{(\sqrt{t})^2}{2!} + \frac{(\sqrt{t})^4}{4!} - \frac{(\sqrt{t})^6}{6!} + ...$

So,

$\displaystyle \frac{\cos(\sqrt{t})}{\sqrt{t}} = \frac{1}{\sqrt{t}} - \frac{(\sqrt{t})^1}{2!} + \frac{(\sqrt{t})^3}{4!} - \frac{(\sqrt{t})^5}{6!} + ...$

$\displaystyle = \sum_{n=0}^{\infty} \frac{(-1)^n (\sqrt{t})^{2n-1}}{(2n)!}$

So the integral of this is...

$\displaystyle \sum_{n=0}^{\infty} \frac{2t(-1)^n (\sqrt{t})^{2n-1}}{(2n+1) (2n)!} + C = \sum_{n=0}^{\infty} \frac{2(-1)^n (\sqrt{t})^{2n+1}}{(2n+1)!} + C$.

Now the power series of $\displaystyle \sin(\sqrt{t})$ is...

$\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n (\sqrt{t})^{2n+1}}{(2n+1)!}$...

So... Can ya finish..?

4. Thanks for taking the time to help me with this Deadstar. I really appreciate this.

Wow, what you posted was an eyefull, unfortunately, I don't understand. Ok, I am new to this, but from my understanding (correct me if I am wrong), I am supposed to use U-substitution, meaning I am supposed to assign U = something, then find du = the derivative of something dx, etc etc. Can you tell me what should I assign U = to?

5. Originally Posted by sparky
Thanks for taking the time to help me with this Deadstar. I really appreciate this.

Wow, what you posted was an eyefull, unfortunately, I don't understand. Ok, I am new to this, but from my understanding (correct me if I am wrong), I am supposed to use U-substitution, meaning I am supposed to assign U = something, then find du = the derivative of something dx, etc etc. Can you tell me what should I assign U = to?
Do you have to solve it that way or is that what you think you have to do. I did not use a u-substitution method.

The way I solved it was to expand $\displaystyle \cos(\sqrt{t})$ as a power series.

See part 3 here. Trigonometric functions - Wikipedia, the free encyclopedia

Then divided it through by $\displaystyle \sqrt{t}$ to get the integral in your question.

Then I used the fact...

The integral of a sum is the sum of an intergal.

This basically just follows from the following property of integrals.

$\displaystyle \int f(x) + g(x) dx = \int f(x) dx + \int g(x) dx$

So all I had to do was integrate the sum formula that generated my series which was...

$\displaystyle = \sum_{n=0}^{\infty} \frac{(-1)^n (\sqrt{t})^{2n-1}}{(2n)!}$

Which is equal to the power series for $\displaystyle \cos(x)$ multiplied by $\displaystyle \frac{1}{\sqrt{t}}$.

I.e. I did this...

$\displaystyle = \int \sum_{n=0}^{\infty} \frac{(-1)^n (\sqrt{t})^{2n-1}}{(2n)!} dt$

= $\displaystyle \sum_{n=0}^{\infty} \int \frac{(-1)^n (\sqrt{t})^{2n-1}}{(2n)!} dt$

= $\displaystyle \sum_{n=0}^{\infty} \frac{2t(-1)^n (\sqrt{t})^{2n-1}}{(2n+1) (2n)!} + C$

(why?)

because...

$\displaystyle \sum_{n=0}^{\infty} \int \frac{(-1)^n (\sqrt{t})^{2n-1}}{(2n)!} dt$

= $\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n }{(2n)!} \int (\sqrt{t})^{2n-1} dt$

= $\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n }{(2n)!} \int t^{1/2(2n-1)} dt$

= $\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n }{(2n)!} \int t^{n-1/2} dt$

= $\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n }{(2n)!} \frac{t^{n+1/2}}{n+1/2} dt$

= $\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n }{(2n)!} \frac{2t^{n+1/2}}{2n+1} dt$

= $\displaystyle \sum_{n=0}^{\infty} \frac{2(-1)^n (\sqrt{t})^{2n+1}}{(2n+1)!} + C$

Then I just realized it was almost the same as the power series for $\displaystyle \sin(\sqrt{t})$ but multiplied by 2.

Have a look at that link and you'll see the series expansions for cos and sin. Just put $\displaystyle \sqrt{t}$ where the x's are and you get my series.

6. ## hi

wow!!!!!!!! its nice but sparky your right with your idea...he just explore it to understand be easily...keep doin that i like calculus so much actually all kinds of math...

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