**Use series to approximate the definite integral $I$. (Round your answer to three decimal places.)
$$I=\int _0^1x \cos\left(x^4\right) \, dx$$
I understand:
$$\int _0^1x \cos(x^2) \, dx=\frac{\sin(1)} 2 \quad \left(\text{Decimal: } 0.42074\ldots \right) = 0.421$$
But when $x$ is to the power of $4$ would it just be:
$$\int _0^1x\cos(x^4) \, dx=\frac{\sin(1)} 4 \quad \left(\text{Decimal: } 0.210 \right)$$
Best Answer
It said "Use series".
You have $$ \cos x = 1 - \frac{x^2} 2 + \frac{x^4}{24} - \frac{x^6}{720} + \cdots. $$ So \begin{align} \cos(x^4) & = 1 - \frac{(x^4)^2} 2 + \frac{(x^4)^4}{24} - \frac{(x^4)^6}{720} + \cdots \\[10pt] & = 1 - \frac{x^8} 2 + \frac{x^{16}}{24} - \frac{x^{24}}{720} + \cdots \end{align} and then $$ x\cos(x^4) = x - \frac{x^9} 2 + \frac{x^{17}}{24} - \frac{x^{25}}{720} + \cdots $$ Integrate that from $0$ to $1$ and notice that after a few terms the alternating positive and negative amounts being added are so small that they cannot change the first three digits after the decimal point.