Evaluate $\int x \cos x^2 dx$ Hay I have hit this in my book 
Evaluate $\int x \cos x^2 dx$.
I got $x^2 \sin(x^2) / 2 $
But I used a online calculator to check it and it is giving me $\sin(x^2)/2 $
Where dose my X go? 
 A: U-substitution allows us to "reverse"/invert the process one uses with the chain rule in differentiating. 
Challenge: Try differentiating your answer, and then differentiating the given solution, and see which one gives you the expression in the integrand, below, in $(1)$?
Now, we have the following integral.
$$\int x \cos(x^2 )dx \tag{1}$$
Using u - substitution:
Let $\color{blue}{\bf u = x^2};\;$ then $\,du = 2x dx \implies \color{red}{x\,dx = \frac 12 du}$
Substitute: $$\int \color{red}{\bf x } \cos(\color{blue}{\bf x^2 })\color{red}{\bf \,dx} \quad = \quad \color{red}{\bf \frac 12} \int \cos \color{blue}{\bf u} \color{red}{\bf\,du} \quad = \quad \frac 12 \sin u + C$$
Back substituting: $$\frac 12 \sin u + C\; =\;\frac 12 \sin(x^2) + C \;= \;\frac{\sin(x^2)}{2} + C$$
A: Let $u=x^2$. Then $du = 2xdx$. Hence
$$
\int x \cos(x^2)\ dx = \frac{1}{2}\int \cos(u)\ du = \frac{\sin(u)}{2} + C = \frac{\sin(x^2)}{2} + C.
$$
A: $\int x \cos(x^2) dx=\int \frac{1}{2}\cos(x^2) d(x^2)=\frac{1}{2}\sin(x^2)+C$
