Compute the integral $$\displaystyle\int\dfrac {x \sin (x^2)}{1+ \cos^2 (x^2)} \ dx$$
Development:
We will use the following procedure:
- Observe what type of almost direct integral the given expression could be. To do so, we have to identify the elements involved in the chain rule.
Although this integral could seem very complicated, we can see in the denominator an element of the form $$1+x^2$$, having $$\cos(x^2)$$ instead of $$x$$.
This should make us think that perhaps we have an arctangent type of integral:
$$$\int\dfrac {x \sin (x^2)}{1+ \cos^2 (x^2)} \ dx=\int\dfrac {x \sin (x^2)}{1+ (\cos(x^2))^2} \ dx$$$
Now then, we know that, $$\dfrac{d}{dx}(\cos(x^2))=-2\cdot x\cdot\sin(x^2)$$, so only we need $$-2$$ in the numerator, so that this is the derivative of $$\cos(x^2)$$, therefore:
$$$\int\dfrac {x \sin (x^2)}{1+ (\cos(x^2))^2} \ dx= -\dfrac{1}{2}\int\dfrac{-2\cdot x \cdot\sin(x^2)}{1+(\cos(x^2))^2} \ dx$$$ and the integral of the arctangent of $$\cos(x^2)$$ times a constant.
- We can use the formula to find the integral: $$$-\dfrac{1}{2}\int\dfrac{-2\cdot x \cdot\sin(x^2)}{1+(\cos(x^2))^2} \ dx= -\dfrac{1}{2}\cdot\arctan(\cos(x^2))+C$$$
Solution:
$$\displaystyle\int\dfrac {x \sin (x^2)}{1+ \cos^2 (x^2)} \ dx=-\dfrac{1}{2}\cdot\arctan(\cos(x^2))+C $$