Evaluate the integral: I = integral (cos x + 1)/(sin x +
Evaluate the integral: $$ I = \int \frac{\cos x + 1}{\sin x + x} \, dx. $$
Explanation of Method
Although the denominator mixes a trig function with a polynomial term, its derivative $\tfrac{d}{dx}(\sin x+x)=\cos x+1$ matches the numerator exactly. Recognizing this pattern is what makes the integral elementary — despite the mixed function types, this is nothing more than the standard logarithmic substitution.
- Substitute $u$ for the denominator.
- Recognize that $du$ exactly matches the numerator.
- Integrate using the standard logarithmic antiderivative.
- Substitute back and add the constant of integration.
Step 1 · Recognize the derivative pattern
We are given the integral:
$$ I = \int \frac{\cos x+1}{\sin x+x}\,dx. $$
Notice that the denominator's derivative is:
$$ \frac{d}{dx}(\sin x+x) = \cos x+1, $$
which matches the numerator exactly. We let:
$$ u = \sin x + x. $$
Then:
$$ du = (\cos x+1)\,dx. $$
Step 2 · Rewrite and integrate
Substituting into the integral:
$$ I = \int \frac{du}{u}. $$
The integral of $\dfrac{1}{u}$ is a standard result:
$$ I = \ln|u| + C. $$
Step 3 · Substitute back
Substituting back $u = \sin x+x$, we obtain:
$$ I = \ln|\sin x+x| + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $u = \sin x + x$ |
| $du$ | $(\cos x+1)\,dx$ — matches the numerator exactly |
| Antiderivative | $\ln|u|$ |
Additional Notes
- The domain of $I$ requires $\sin x+x \neq 0$, which holds for all $x$ except $x=0$ (the only real solution to $\sin x=-x$).
- Despite mixing a trig function and a polynomial, the denominator's derivative matching the numerator is the same signal used in Problems 67, 78, and 79 — always worth checking before reaching for more advanced techniques.
- The result can be checked by differentiating $\ln|\sin x+x|$ with the chain rule and confirming it returns $\dfrac{\cos x+1}{\sin x+x}$.
Result
The solution to the integral is: