Evaluate the integral: I = integral (sin x)/(sqrt(2+ cos
Evaluate the integral: $$ I = \int \frac{\sin x}{\sqrt{2+\cos x}} \, dx. $$
Explanation of Method
The derivative of $2+\cos x$ is $-\sin x$, which is (up to sign) exactly the numerator. This makes $u=2+\cos x$ the natural substitution, reducing the integral to a simple power-rule form $\int u^{-1/2}\,du$.
- Substitute $u=2+\cos x$.
- Rewrite the integral in terms of $u$.
- Apply the power rule for integration.
- Substitute back and add the constant of integration.
Step 1 · Set up the substitution
We are given the integral:
$$ I = \int \frac{\sin x}{\sqrt{2+\cos x}}\,dx. $$
Let:
$$ u = 2+\cos x. $$
Then:
$$ du = -\sin x\,dx \quad \Longrightarrow \quad \sin x\,dx = -du. $$
Step 2 · Rewrite the integral
Substituting $u$ and $-du$ into the integral:
$$ I = \int \frac{-du}{\sqrt{u}} = -\int u^{-1/2}\,du. $$
Step 3 · Apply the power rule
Using the power rule for integration, $\displaystyle\int u^n\,du = \frac{u^{n+1}}{n+1}$ with $n=-\tfrac12$:
$$ -\int u^{-1/2}\,du = -\frac{u^{1/2}}{1/2} + C = -2\sqrt{u} + C. $$
Step 4 · Substitute back
Substituting back $u = 2+\cos x$, we obtain:
$$ I = -2\sqrt{2+\cos x} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $u = 2+\cos x$ |
| $\sin x\,dx$ in terms of $du$ | $-du$ |
| Antiderivative | $-2\sqrt{u}$ |
Additional Notes
- The integrand is defined for all real $x$, since $2+\cos x \geq 1 > 0$ always, so the square root and the division are never problematic.
- This is the mirror image of Problem 92's pattern: whenever the numerator matches (up to sign) the derivative of the expression under a radical in the denominator, the substitution collapses the integral to a plain power-rule form.
- The result can be checked by differentiating $-2\sqrt{2+\cos x}$ with the chain rule and confirming it returns $\dfrac{\sin x}{\sqrt{2+\cos x}}$.
Result
The solution to the integral is: