Evaluate $\int x \cos x^2 dx$

Question

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?

Answer 1

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: $$\frac 12 \sin u + C\; =\;\frac 12 \sin(x^2) + C \;= \;\frac{\sin(x^2)}{2} + C$$

Answer 2

Let $u=x^2$. Then $du = 2xdx$. Hence $$ \int x \cos(x^2)\ dx = \frac{1}{2}\int \cos(u)\ du = \frac{\sin(u)}{2} + C = \frac{\sin(x^2)}{2} + C. $$

Answer 3

$\int x \cos(x^2) dx=\int \frac{1}{2}\cos(x^2) d(x^2)=\frac{1}{2}\sin(x^2)+C$