Explain each step to find $ \int \frac{\cos x\,d x}{1 + (\sin x)^{2}} $

Question

I know the answer but I don't understand the steps to integrate.

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

Answer 1

$$I= \int \frac{\cos x \,dx }{1 + \sin^2 x} $$

substitute $\sin x=t\iff\cos x \,dx=dt$

$$I= \int \frac{dt }{1 + t^{2}} $$ If this seems obvious then we're done

else substitute $t=\tan\theta\iff dt=\sec^2\theta \,d\theta$ $$I= \int \frac{\sec^2\theta \,d\theta }{1 + \tan^{2}\theta}= \int \frac{\sec^2\theta \,d\theta }{\sec^{2}\theta}=\int\,d\theta=\theta+ C $$ By putting Everything back we get

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

Answer 2

Note that $\mathrm d\Big(f(x)\Big) = f'(x)\ \mathrm dx $ , we can avoid substitutions for simple integrals such as these: $$\require{cancel} \int \frac{\cancel{\cos x} }{1 + \sin^2 x} \frac{\mathrm d\left(\sin x\right)}{\cancel{\cos x}} = \int \frac{\mathrm d(\sin x)}{1 + (\sin x)^2}$$

Now, we know that $\frac{\mathrm d}{\mathrm dx}\arctan x = 1/(1+x^2)$

Obviously, $$\int \frac{\cos x}{ 1+ \sin^2 x}\ \mathrm dx = \arctan\left(\sin x\right) \color{grey}{+ \mathcal C}$$