Question

\[\int_0^\frac{\pi}{4} \frac{\sin x + \cos x}{3 + \sin2x}dx\]

Answer 1

Let

\[I = \int_0^\frac{\pi}{4} \frac{\sin x + \cos x}{3 + \sin2x}dx\]

\[= \int_0^\frac{\pi}{4} \frac{\sin x + \cos x}{4 - \left( 1 - \sin2x \right)}dx\]

\[ = \int_0^\frac{\pi}{4} \frac{\sin x + \cos x}{4 - \left( \sin^2 x + \cos^2 x - 2\sin x\cos x \right)}dx\]
\[ = \int_0^\frac{\pi}{4} \frac{\sin x + \cos x}{4 - \left( \sin x - \cos x \right)^2}dx\]

Put

\[\sin x - \cos x = z\]

\[\therefore \left( \cos x + \sin x \right)dx = dz\]

When

\[x \to 0, z \to - 1 .................\left( z = \sin0 - \cos0 = 0 - 1 = - 1 \right)\]

When

\[x \to \frac{\pi}{4}, z \to 0 .......................\left( z = \sin\frac{\pi}{4} - \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0 \right)\]

\[\therefore I = \int_{- 1}^0 \frac{dz}{2^2 - z^2}\]
\[ = \left.\frac{1}{2 \times 2}\log\left( \frac{2 + z}{2 - z} \right)\right|_{- 1}^0 \]
\[ = \frac{1}{4}\left( \log1 - \log\frac{1}{3} \right)\]
\[ = \frac{1}{4}\left[ 0 - \left( \log1 - \log3 \right) \right]\]
\[ = - \frac{1}{4}\left( 0 - \log3 \right)\]
\[ = \frac{1}{4}\log3\]