Evaluate the integral: integral sqrt(1 - sin 2x) dx
Evaluate the integral: $$ \int \sqrt{1 - \sin 2x} \, dx. $$
Explanation of Method
The double-angle identity $\sin 2x = 2\sin x\cos x$ lets $1-\sin 2x$ be rewritten as $(\cos x-\sin x)^2$, matching the pattern $a^2-2ab+b^2=(a-b)^2$. Once the expression under the root is a perfect square, the radical simplifies to an absolute value, which (taking the principal branch) reduces the integral to two standard trig antiderivatives.
- Rewrite $\sin 2x$ using the double-angle identity.
- Recognize the resulting expression as a perfect square.
- Simplify the square root and integrate the remaining trig terms.
- Combine the results and add the constant of integration.
Step 1 · Simplify $\sin 2x$ using a trigonometric identity
Using the double-angle identity for sine:
$$ \sin 2x = 2\sin x\cos x, $$
we substitute this expression into the integral. The integral becomes:
$$ \int \sqrt{1 - 2\sin x\cos x}\,dx. $$
To simplify the expression further, we use the identity:
$$ 1 - \sin 2x = 1 - 2\sin x\cos x = (\cos x - \sin x)^2. $$
Thus, the integral becomes:
$$ \int \sqrt{1-\sin 2x}\,dx = \int \sqrt{(\cos x - \sin x)^2}\,dx. $$
Step 2 · Simplify the square root
The square root of $(\cos x-\sin x)^2$ is $|\cos x - \sin x|$. For simplicity, we consider the principal value:
$$ \sqrt{(\cos x - \sin x)^2} = \cos x - \sin x. $$
The integral now becomes:
$$ \int \sqrt{1-\sin 2x}\,dx = \int (\cos x - \sin x)\,dx. $$
Step 3 · Evaluate the integral
We split the integral into two simpler parts:
$$ \int (\cos x - \sin x)\,dx = \int \cos x\,dx - \int \sin x\,dx. $$
First term: $\displaystyle \int \cos x\,dx$. The integral of $\cos x$ is:
$$ \int \cos x\,dx = \sin x. $$
Second term: $\displaystyle \int \sin x\,dx$. The integral of $\sin x$ is:
$$ \int \sin x\,dx = -\cos x. $$
Combining these results:
$$ \int (\cos x - \sin x)\,dx = \sin x + \cos x. $$
Step 4 · Add the constant of integration
Finally, the solution to the integral is:
$$ \int \sqrt{1-\sin 2x}\,dx = \sin x + \cos x + C, $$
where $C$ is the constant of integration.
Summary Table
| Term | Antiderivative |
|---|---|
| $\cos x$ | $\sin x$ |
| $-\sin x$ | $\cos x$ |
Additional Notes
- Strictly, $\sqrt{(\cos x-\sin x)^2}=|\cos x-\sin x|$, so the clean form $\cos x-\sin x$ used above is valid only where $\cos x \geq \sin x$; on other intervals the sign flips and the antiderivative becomes $-\sin x-\cos x+C$.
- The domain of $I$ is all real $x$, since $1-\sin 2x \geq 0$ always (as $\sin 2x \leq 1$).
- The result can be checked by differentiating $\sin x+\cos x$ and confirming it returns $\cos x-\sin x$, the branch of the integrand used here.
Result
The solution to the integral is: