Question

Evaluate the integral: \[ \int_{0}^{\frac{\pi}{4}} \sqrt{1 + \sin^2 x} \, dx \]

Answer 1

Step 1: Trigonometric Identity Transformation

We know the identity:

\[ \sin 2x = 2 \sin x \cos x \] 

Using this in the integrand:

\[ \sqrt{1 + \sin 2x} = \sqrt{1 + 2 \sin x \cos x} \]

Also, recall that:

\[ 1 + \sin 2x = (\sin x + \cos x)^2 \]

So:

\[ \sqrt{1 + \sin 2x} = |\sin x + \cos x| \]

In the interval \( [0, \frac{\pi}{4}] \), both \( \sin x \) and \( \cos x \) are positive, so the absolute value is not needed:

\[ \sqrt{1 + \sin 2x} = \sin x + \cos x \]

Step 2: Integrating the Expression

Now the integral becomes:

\[ I = \int_0^{\frac{\pi}{4}} (\sin x + \cos x) \, dx \]

Split the integral:

\[ I = \int_0^{\frac{\pi}{4}} \sin x \, dx + \int_0^{\frac{\pi}{4}} \cos x \, dx \]

Compute each part:

• \( \int \sin x \, dx = -\cos x \)
• \( \int \cos x \, dx = \sin x \)

So:

\[ I = [-\cos x]_0^{\frac{\pi}{4}} + [\sin x]_0^{\frac{\pi}{4}} \]

Evaluate:

\[ I = (-\cos \frac{\pi}{4} + \cos 0) + (\sin \frac{\pi}{4} - \sin 0) \]

\[ I = \left( -\frac{1}{\sqrt{2}} + 1 \right) + \left( \frac{1}{\sqrt{2}} - 0 \right) \]

\[ I = \left(1 - \frac{1}{\sqrt{2}}\right) + \frac{1}{\sqrt{2}} = 1 \]

Step 3: Conclusion

The value of the integral is:

\[ \int_0^{\frac{\pi}{4}} \sqrt{1 + \sin 2x} \, dx = \boxed{1} \]