The integral 80 0 4 ( sin + cos 9 + 16 sin 2 ) d is equal to : [2025]

Question

The integral $80 \int_{0}^{\frac{π}{4}} (\frac{\sin θ + \cos θ}{9 + 16 \sin 2 θ}) d θ$ is equal to : [2025]

Smart Achievers · Accepted Answer

(1)

Let $I = 80 \int_{0}^{\frac{π}{4}} \frac{(\sin θ + \cos θ) d θ}{(9 + 16 \sin 2 θ)}$

$= 80 \int_{0}^{\frac{π}{4}} \frac{(\sin θ + \cos θ) d θ}{9 + 16 (1 - 1 + 2 \sin θ \cos θ)}$

$= 80 \int_{0}^{\frac{π}{4}} \frac{(\sin θ + \cos θ) d θ}{25 - 16 {(\sin θ - \cos θ)}^{2}}$

Put $\sin θ - \cos θ = t \Rightarrow (\cos θ + \sin θ) d θ = d t$

When $θ = 0, t = - 1 and θ = π / 4, t = 0$ .

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