The integral 80 \int 4 sin \theta+cos \theta ) ( 9+16 d\theta is equal to : 0 sin 2\theta
Step-by-Step Solution
Key Concept: Apply the core result for definite integral properties and simplify using the given constraints.
4 sin \theta + cos \theta I = \int ( ) d\theta (2) 0 9 - 16 sin 2\theta Take sin \theta - cos \theta = t (cos \theta + sin \theta)d\theta = dt 2 2 (sin \theta - cos \theta) = t 2 \Rightarrow sin 2\theta = 1 - t \theta = 0 \to t = -1 \pi \theta = \to t = 0 4 0 dt I = \int 2 -1 9 + 16 (1 - t ) 0 1 dt 16 2 = \int - t 16 -1 25 16 0 1 1 5-4t = [ log 5+4t ] ∣ 4 10 ∣ -1 1 = [0 + log 9] e 40 log 9 e I = 40 80I = 2 log 9 e 80I = 4 log 3 e
Correct Answer: 2