Answer to Question #234229 in Calculus for Sayem

Question #234229

If f is continuous and f(x)dx=6(upperlimit=2,lowerlimit=0)\intop f (x) dx=6 (upper limit=2, lower limit=0) ,evaluate

f(2sinθ)cosθdθ(upperlimit=π/2,lowerlimit=0)\intop f(2sin\theta) cos\theta d\theta (upper limit=\pi/2 ,lower limit=0)


1
Expert's answer
2021-09-07T19:00:15-0400

Solution:

Given,02f(x)dx=6\int_0^2 f(x)dx=6 ...(i)

Let I=0π/2f(2sinθ)cosθdθI=\int_0^{\pi/2} f(2\sin \theta)\cos \theta d\theta

Put 2sinθ=t2\sin \theta=t

2cosθ=dtdθ\Rightarrow 2\cos \theta=\dfrac{dt}{d\theta}

When θ=0,t=0;θ=π/2,t=2\theta=0,t=0; \theta=\pi/2,t=2

So, I=1202f(t)dtI=\frac12\int_0^{2}f(t)dt

=12×6=\frac12\times 6 [Using (i)]

=3=3


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