Q.Evaluate the definite integral $\int_{\pi/6}^{\pi/3}\frac{\sin x+\cos x}{\sqrt{\sin 2x}}\,dx$

Write $I=\displaystyle\int_{\pi/6}^{\pi/3}\frac{\sin x+\cos x}{\sqrt{\sin 2x}}\,dx$ and put $u=\sin x-\cos x$, so $du=(\cos x+\sin x)\,dx$ — precisely the numerator.

Since $(\sin x-\cos x)^2=1-2\sin x\cos x=1-\sin 2x$, we have $\sin 2x=1-u^2$. The limits move to $u=\tfrac{1-\sqrt3}{2}$ (at $x=\tfrac\pi6$) and $u=\tfrac{\sqrt3-1}{2}$ (at $x=\tfrac\pi3$): …

1. Set up the integral and choose the substitution. Let

Differentiating, $\dfrac{du}{dx}=\cos x+\sin x$, so $du=(\sin x+\cos x)\,dx$ — this matches the numerator exactly.

2. Rewrite $\sin 2x$ through $u$. Squaring $u$,

because $\sin^2x+\cos^2x=1$ and $2\sin x\cos x=\sin 2x$. Therefore

3. Convert the limits.

   The substitution replaces $x$-limits by $u$-limits:

   • at $x=\dfrac{\pi}{6}$: $u=\sin\dfrac{\pi}{6}-\cos\dfrac{\pi}{6}=\dfrac12-\dfrac{\sqrt3}{2}=\dfrac{1-\sqrt3}{2}$;
   • at $x=\dfrac{\pi}{3}$: $u=\sin\dfrac{\pi}{3}-\cos\dfrac{\pi}{3}=\dfrac{\sqrt3}{2}-\dfrac12=\dfrac{\sqrt3-1}{2}.$

   Notice the two limits are negatives of each other: $\dfrac{1-\sqrt3}{2}=-\dfrac{\sqrt3-1}{2}$.

4. Integrate in $u$.

   The integral becomes the standard arcsine form: …

Method: Substitute $u=\sin x-\cos x$ when the numerator is $\sin x+\cos x$

Use this for integrands with $\sin x+\cos x$ on top and a $\sin 2x$-dependent denominator: $u=\sin x-\cos x$ makes the numerator exactly $du$ and links the denominator via $u^2=1-\sin 2x$.

Steps

Step 1: Choose $u=\sin x-\cos x$.

Then $du=(\cos x+\sin x)\,dx$, matching the numerator.

Step 2: Rewrite the denominator through $u$.

Square: $u^2=1-2\sin x\cos x=1-\sin 2x$, so $\sin 2x=1-u^2$ and $\sqrt{\sin 2x}=\sqrt{1-u^2}$. …

Common Mistakes

Mistake 1: Not seeing $u^2=1-\sin 2x$.

Why it's wrong: the link $\sqrt{\sin 2x}=\sqrt{1-u^2}$ is what turns the integral into an arcsine; missing it leaves a hard radical. Correct approach: square $u=\sin x-\cos x$.

Mistake 2: Forcing a "nice" angle for $2\arcsin\!\bigl(\tfrac{\sqrt3-1}{2}\bigr)$.

Why it's wrong: $\tfrac{\sqrt3-1}{2}$ is not a standard sine value, so the exact answer is best left in arcsine form. Correct approach: accept the exact expression. …