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Question
Evaluate the definite integral:
`int_0^(pi/4) (sin x + cos x)/(9+16sin 2x) dx`
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Solution
Let `I = int_0^(pi/4) (sin x + cos x)/(9 + 16 sin 2x)`dx
Put sin x - cos x = t
(cos x + sin x)dx = dt
and 1 - 2 sin x cos x = t2
⇒ sin 2x = 1 - t2
When x = `pi/4`, t = sin `pi/4 - cos pi/4`
`= 1/sqrt2 - 1/sqrt2 = 0`
When x = 0, t = sin 0 - cos 0 = - 1
`therefore int_0^(pi/4) (sin x + cos x)/(9 + 16 sin 2x)`dx
`= int_(- 1)^0 dt/(9 + 16 (1 - t^2))`
`= int_(- 1)^0 dt/(25 - 16 t^2)`
`= 1/16 int_(- 1)^0 dt/((5/4)^2 - t^2)`
`= 1/16 * 1/(2 * 5/4) [log |(5/4 + t)/(5/4 - t)|]_(-1)^0`
`= 1/40 [log 1 - (log 1 - log 9)]`
`= 1/40 log 9`
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