Question

If f is continuous and interval[0,2] f(x) dx =6, find interval [0,pi/2] 3f(2sinx)cosx dx

Answer 1

I assume by that you mean integral, not interval, right?

Answer 2

well i mean the upper limit is 2 and lower is 0 like \[\int\limits_{0}^{2}f
(x)dx\] , find \[\int\limits_{0}^{\pi/2} 3f(2sinx)cosxdx\]

Answer 3

So, yes, you mean integral.

Answer 4

Yeah, that's what I thought. Let
\[x = 2sin(\theta) \rightarrow dx = 2cos(\theta)d\theta\]
then we have that
\[\theta = \sin^{-1}(\frac{x}{2})\]
so at the endpoints,
\[\theta = \sin^{-1}(0) = 0, \theta = \sin^{-1}(1) = \frac{\pi}{2} \]
so the integral becomes
\[\int_0^{\frac{\pi}{2}} f(2\sin(\theta))\cdot 2\cos(\theta) d\theta\]
The substitution doesn't change the value of the integral, so we have that
\[2\int_0^{\frac{\pi}{2}} f(2\sin(\theta))\cdot\cos(\theta) d\theta = 6\]
so
\[\int_0^{\frac{\pi}{2}} f(2\sin(\theta))\cdot\cos(\theta) d\theta = 3 \]
The question asks for
\[3\int_0^{\frac{\pi}{2}} f(2\sin(\theta))\cdot\cos(\theta) d\theta \]
which is apparently 3*3 = 9.