How to solve the antiderivative of $x\cos\left(x^3\right)$ What is the way to solve:
$$\int x\cos\left(x^3\right)\space\text{d}x$$
Thanks, I've no idea how to start
 A: 
The Maclaurin series of a function:

*

*$$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)x^n}{n!}$$
Notice, the Maclaurin series of $\cos(x)$:

*

*$$\cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}$$
So, we can find the sum when $x^3$:

*

*$$\cos\left(x^3\right)=\sum_{n=0}^{\infty}\frac{(-1)^n\left(x^3\right)^{2n}}{(2n)!}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{6n}}{(2n)!}$$
Now, we can find the sum when we multiply both sides by $x$:

*

*$$x\cos\left(x^3\right)=x\sum_{n=0}^{\infty}\frac{(-1)^nx^{6n}}{(2n)!}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{6n+1}}{(2n)!}$$


Now, we can find the integral, using $\int s^b\space\text{d}s=\frac{s^{b+1}}{b+1}+\text{C}$:
$$\int x\cos\left(x^3\right)\space\text{d}x=\int\sum_{n=0}^{\infty}\frac{(-1)^nx^{6n+1}}{(2n)!}\space\text{d}x=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}\int x^{6n+1}\space\text{d}x=$$
$$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}\cdot\frac{x^{6n+2}}{6n+2}+\text{C}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{6n+2}}{(6n+2)(2n)!}+\text{C}$$
A: $\quad\displaystyle\int x\cos(x^3)\mathrm dx$
$=\displaystyle\int x\sum_{k\mathop=0}^\infty\frac{(-1)^k(x^3)^{2k}}{(2k)!}\mathrm dx$
$=\displaystyle\int\left(\sum_{k\mathop=0}^\infty\frac{(-1)^kx^{6k+1}}{(2k)!}\right)\mathrm dx$
$=\displaystyle\sum_{k\mathop=0}^\infty\int\frac{(-1)^kx^{6k+1}}{(2k)!}\mathrm dx$
$=\displaystyle\sum_{k\mathop=0}^\infty\frac{(-1)^kx^{6k+2}}{(2k)!(6k+2)}+C$
A: You can first write $\cos(x^3)$ using exponentials and apply linearity: $$\int \frac{x\left(e^{ix^3} + \frac{1}{e^{ix^3}}\right)}{2} dx = \frac 12 \left(\int x\left(e^{ix^3} + \frac{1}{e^{ix^3}}\right) dx\right)$$
Solvivng the Integral
First, we need to cancel and/or combine common factors (in case here, we have $e^{ix^3}$): $$\int \left(\frac{x\left(e^{2ix^3} + 1\right)}{e^{ix^3}} dx\right)$$ Then, we can expand the integral and apply linearity: $$\int xe^{ix^3} + \frac{x}{e^{ix^3}} dx = \left(\int xe^{ix^3} dx\right) + \left(\int \frac{x}{e^{ix^3}} dx\right)$$
Solving Part I
To solve the first integral, we can substitute $-x^2 \to u$, which means that $\frac{du}{dx} = -2x$: $$-\frac 12\left(\int \frac{1}{e^{u^{1/2}}} du\right)$$ The integral simplifies to: $$-\frac{2\Gamma(\frac 23, u^{3/2})}{3}$$ When we multiply that by $\frac 12$, we get: $$\frac{\Gamma(\frac 23, u^{3/2})}{3}$$ We can undo the substitution and get: $$\frac{\Gamma(\frac 23, -ix^3)}{3}$$
Solving Part II
We apply the same method we did in part I, and we will get the same answer as before EXCEPT that the second parameter in the gamma function is actually $\lvert -ix^3 \rvert$.
Final Part
When we plug in the integrals, we get: $$\frac{\Gamma(\frac 23, ix^3)}{3} + \frac{\Gamma(\frac 23, -ix^3)}{3}$$ Then, we multiply both addends by $\frac 12$ to get: $$\frac{\Gamma(\frac 23, ix^3)}{6} + \frac{\Gamma(\frac 23, -ix^3)}{6}$$ We finally simplify that to: $$\frac{\Gamma(\frac 23, ix^3) + \Gamma(\frac 23, -ix^3)}{6}$$
So, that is your final answer.
