I'm an adult learner, slowly working my way through James Stewart's Calculus 8th Edition - Single Variable Calculus Early Transcendentals.
I am currently working through some exercises on indefinite integrals, specifically problems pertaining to the Substitution Rule (Section 5.5), and am at a loss to understand the justification for a particular inference provided in the solutions manual I have available to me.
The Problem
Find the indefinite integral of the following:
$$ \int{\frac{\sin{2x}}{1 + \cos^2 x}}\ dx $$
which is transformed using the double angle formula to:
$$ 2 \int{\frac{\sin x \cos x}{1 + \cos^2 x}}\ dx $$
The Approach
Letting $u = \cos x $,
$$ \begin{align} \frac{du}{dx} &= -\sin x\ \\ du &= -\sin x\ dx \\ \frac{du}{- \sin x} &= dx \end{align} $$
Therefore, the original problem may be restated using the Substitution Rule as follows:
$$ \begin{align} 2 \int{\frac{\sin x\ u}{1+u^2}\ \frac{-du}{\sin x}} & = -2 \int{\frac{u}{1+u^2}}\ du \\ \end{align} $$
The Troublesome Inference
At this juncture, I am at a loss of what to do. To date, this stage of these exercises required inverting some one of the rules of differentiation, but I feel blind as to which I am meant to use here.
The solutions manual provides the following inference:
$$ \begin{align} 2 \int{\frac{\sin x\ u}{1+u^2}\ \frac{-du}{\sin x}} & = -2 \int{\frac{u}{1+u^2}}\ du \\ & = -2 \cdot \frac{1}{2} \ln{1 + u^2} + C \end{align} $$
The Question
While I do understand that $\frac{d}{dx} \ln{x} = \frac{1}{x}$, wouldn't an antiderivative of $\frac{-1}{1 + u^2}$ be some composite function requiring the Chain Rule to differentiate?
I'm struggling to understand the inference based on the tools I currently have available to me.
Hopefully this is clear enough that someone can provide some guidance.
Thank you,
Pierre
Find the indefinite integral of the following:
$$ \int{\frac{\sin{2x}}{1 + \cos^2 x}}\ dx $$
Step 1 :
$$ 2 \int{\frac{\sin x\space{cos ~x}}{1 + \cos^2 x}}\ dx $$
Step 2 : use-base substitution
Let $u = cos~x~,$
$$ \frac{du}{dx}=-~sin~x $$
$$ =-2 \int{\frac{u}{1 + u^2 }}\ du $$
Step 3 : again base substitution
Let $v = 1 + u^2 ~, $
$$\frac{du}{dx}= 1 + 2~u$$
$$= -\int{\frac{1}{v }}\ dv $$
$$= -~ln|v|+c $$
Final step :
$$ =-~ln|~{1 + \cos^2 x}| + c $$