Question

Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .

Answer 1

Let

\[I = \int_0^{{\pi}/{4}}\frac{\sin x + \cos x}{16 + 9\sin2x}dx\]

Put – cosx + sinx = t    .....(1)
Then,
(sin x + cos x) dx = dt
As, x → 0, t → −1
Also, x → \[\frac{\pi}{4}\] t → 0

Squaring (1) both sides, we get
cos²x + sin²x – 2cosx sinx = t²
⇒ 1 – sin2x = t²
⇒ sin 2x = 1 – t²
Substituting these values, we get

\[I = \int_{- 1}^0 \frac{dt}{16 + 9 \left( 1 - t^2 \right)}\]

\[ = \int_{- 1}^0 \frac{dt}{25 - 9 t^2}\]

\[ = \frac{1}{9} \int_{- 1}^0 \frac{dt}{\left( \frac{5}{3} \right)^2 - t^2}\]

\[ = \frac{1}{9} \left[ \frac{1}{2a}\log \left| \frac{a + t}{a - t} \right| \right]_{- 1}^0 \text { where a } = \frac{5}{3}\]

\[ = \frac{1}{9} \left[ \frac{3}{2\left( 5 \right)}\log \left| \frac{\frac{5}{3} + t}{\frac{5}{3} - t} \right| \right]_{- 1}^0 \]

\[ = \frac{1}{9} \left[ \frac{3}{10}\left\{ \log 1 - \log \frac{1}{4} \right\} \right]^{- 1} \]

\[ = \frac{3}{90}\left( - \log \frac{1}{4} \right) = \frac{1}{30} \log 4\]