How to find MacLaurin polynomial of high degree

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. $${\int_0^{0.68}\sin(5x^2)dx}$$

• For this function it is trivial to find the whole Maclaurin series. Please expand $\sin(5t^2)$ first. Use the standard sine Maclaurin expansion and a trivial substitution. Next integrate this series. – szw1710 Sep 21 '18 at 21:38

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

so that

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

• I'm not sure it's such a good idea to spoon-feed the answer to an OP who's posted so many low-effort questions – Jam Sep 21 '18 at 21:45
• @Jam, I agree with you totally, but my own natural reaction in a case like this is to rush in to solve the problem for the OP, and only afterwards to reflect that I probably did to much of the work for them. – Lubin Sep 21 '18 at 21:49
• @Jam: " who's posted so many low-effort questions": how do I know ? – Yves Daoust Sep 22 '18 at 8:16

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.

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