Maclaurin Series for ##\int_{0}^{x} \cos{t^2} \cdot dt##

[Potatochip911]

Homework Statement

Find the Maclaurin series of $\int_{0}^{x} \cos{t^2} \cdot dt$

Homework Equations

3. The Attempt at a Solution [/B]
I normally have some idea how to go about solving these but for this one I just can't figure out where to start. I tried doing it with $\int_{0}^{x} \cos{t}\cdot dt$ and then after evaluating that series replacing $x$ with $x^2$ but that didn't work at all. I checked what the integral of the original function was on wolfram alpha and it looked quite complicated so I'm assuming I'm supposed to calculate the integral of a simpler function then replace the x but I can't figure out what the simpler function is.

 

[Mark44]

Homework Statement

Find the Maclaurin series of $\int_{0}^{x} \cos{t^2} \cdot dt$

Homework Equations

3. The Attempt at a Solution [/B]
I normally have some idea how to go about solving these but for this one I just can't figure out where to start. I tried doing it with $\int_{0}^{x} \cos{t}\cdot dt$ and then after evaluating that series replacing $x$ with $x^2$ but that didn't work at all. I checked what the integral of the original function was on wolfram alpha and it looked quite complicated so I'm assuming I'm supposed to calculate the integral of a simpler function then replace the x but I can't figure out what the simpler function is.

Write the Maclaurin series for cos(t²), then integrate that series term-by-term to get the Maclaurin series of the integral you posted. Finally, evaluate your series at x and at 0, just like you would normally do with a definite integral.

 

[Potatochip911]

Write the Maclaurin series for cos(t²), then integrate that series term-by-term to get the Maclaurin series of the integral you posted. Finally, evaluate your series at x and at 0, just like you would normally do with a definite integral.

Okay so $$\cos{t}=1-\frac{1}{2!}t^2+\frac{1}{4!}t^4-\frac{1}{6!}t^6+...\\
\cos{t^2}=1-\frac{1}{2!}t^4+\frac{1}{4!}t^8-\frac{1}{6!}t^{12}+... \\
\int_{0}^{x} \cos{t^2}\cdot dt=\int_{0}^{x} (1-\frac{1}{2!}t^4+\frac{1}{4!}t^8-\frac{1}{6!}t^{12}+...)dt \\
\int_{0}^{x} \cos{t^2} \cdot dt=x-\frac{1}{5*2!}x^5+\frac{1}{9*4!}x^9-\frac{1}{12*6!}x^{13}+... \\
\int_{0}^{x} \cos{t^2} \cdot dt=\sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+1}}{(4n+1)!(2n)!} $$
Thanks this ended up giving the correct answer.

 

[Ray Vickson]

Okay so $$\cos{t}=1-\frac{1}{2!}t^2+\frac{1}{4!}t^4-\frac{1}{6!}t^6+...\\
\cos{t^2}=1-\frac{1}{2!}t^4+\frac{1}{4!}t^8-\frac{1}{6!}t^{12}+... \\
\int_{0}^{x} \cos{t^2}\cdot dt=\int_{0}^{x} (1-\frac{1}{2!}t^4+\frac{1}{4!}t^8-\frac{1}{6!}t^{12}+...)dt \\
\int_{0}^{x} \cos{t^2} \cdot dt=x-\frac{1}{5*2!}x^5+\frac{1}{9*4!}x^9-\frac{1}{12*6!}x^{13}+... \\
\int_{0}^{x} \cos{t^2} \cdot dt=\sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+1}}{(4n+1)!(2n)!} $$
Thanks this ended up giving the correct answer.

Alternatively: if $f(x) = \int_0^x \cos (t^2) \, dt$ the Maclauren series is
$$f(0) + f'(0) x + \frac{1}{2!} f''(0) x^2 + \cdots,$$
whose terms are easy to get, since
$$\begin{array}{rcl}<br /> f(0) &=& \int_0^0 \cos (t^2) \, dt = 0\\<br /> f'(x) &=& \cos (x^2) \; \Rightarrow f'(0) = \cos^2(0) = 1 \\<br /> f''(x) &=& \frac{d}{dx} \cos(x^2) = -2 x \sin(x^2) \; \Rightarrow f''(0) = 0 \\<br /> &\vdots&<br /> \end{array}$$
In fact, $f'(0), f''(0), \ldots$ are just the coefficients in the Maclauren expansion of $\cos(x^2)$, multiplied by the appropriate factorial.