How to find MacLaurin polynomial of high degree

Question

Given the function $${F(x)=\int_0^x\sin(5t^2)dt}$$

I must find the MacLaurin polynomial of degree $7$ for $F(x)$. Given as a function of $x$.

Since we already have the first derivative given by the integral, I continued to integrate until I reached a $7$th degree within the function and plugged in. This seems to be clearly wrong and very time consuming. How would I proceed to find the MacLaurin polynomial.

After, I'm asked to find the the value of the integral given below using the polynomial found above. $${\int_0^{0.68}\sin(5x^2)dx}$$

Answer 1

$$\sin(5t^2)\approx 5t^2-\frac{(5t^2)^3}{3!}+\frac{(5t^2)^5}{5!}-\frac{(5t^2)^7}{7!}+\cdots$$

so that

$$\int_0^x\sin(5t^2)dt\approx\frac{5x^3}3-\frac{5^3x^7}{7\cdot3!}+\frac{5^5x^{11}}{11\cdot5!}-\frac{5^7x^{15}}{15\cdot7!}+\cdots$$

Answer 2

Hint: I'm guessing the Maclaurin polynomial of degree $7$ is the Taylor series at $0$ (up to degree $7$)... So, using Taylor's theorem: $f(x)=\sum_{n=0}^{\infty}\frac{f^n(0)}{n!}x^n$ we plug in the first seven derivatives of $f(x)$ at $0$.

The fundamental theorem of calculus (FTC) gives us the first: $f'(0)=\sin(5\cdot0^2)=\sin0=0$.

Now differentiate $6$ more times (and evaluate at $0$). Start with $f'(x)=\sin(5x^2)$...

To do the second part plug $x=0.68$ into the polynomial.

Answer 3

$$F(x)=\int_0^x\sin(5t^2)dt=\int\limits_0^x\sum\limits_{k=0}^\infty\frac{(-1)^k5^{2k+1}}{(2k+1)!}t^{4k+2}dt$$$$ =\sum_{k=0}^\infty\frac{(-1)^k5^{2k+1}}{(4k+3)(2k+1)!}x^{4k+3}$$ and since it is up to polynomial of order $7$ we will only need the first two terms, so: $$F(x)\approx\frac{(-1)^05^1}{(3)*1!}x^3+\frac{(-1)^15^3}{(7)*3!}x^7$$$$ =\frac{5}{3}x^3+\frac{125}{42}x^7$$