Answer
$\int_{0}^{\pi/2} f(2sin~\theta)~cos~\theta~d\theta = 3$
Work Step by Step
The function $f$ is continuous.
Let $g(\theta) = 2sin~\theta$
Then $g'(\theta) = 2cos~\theta$
$g'(\theta)$ is continuous for all values of $\theta$
According to the Substitution Rule:
$\int_{0}^{\pi/2} f(g(\theta))~g'(\theta)~d\theta = \int_{g(0)}^{g(\pi/2)} f(x)~dx$
$\int_{0}^{\pi/2} f(2sin~\theta)~(2cos~\theta)~d\theta = \int_{0}^{2} f(x)~dx$
$2\cdot \int_{0}^{\pi/2} f(2sin~\theta)~cos~\theta~d\theta = 6$
$\int_{0}^{\pi/2} f(2sin~\theta)~cos~\theta~d\theta = 3$
