If f is continuous and ∫f(x)dx=6(upperlimit=2,lowerlimit=0)\intop f (x) dx=6 (upper limit=2, lower limit=0)∫f(x)dx=6(upperlimit=2,lowerlimit=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)∫f(2sinθ)cosθdθ(upperlimit=π/2,lowerlimit=0)
Solution:
Given,∫02f(x)dx=6\int_0^2 f(x)dx=6∫02f(x)dx=6 ...(i)
Let I=∫0π/2f(2sinθ)cosθdθI=\int_0^{\pi/2} f(2\sin \theta)\cos \theta d\thetaI=∫0π/2f(2sinθ)cosθdθ
Put 2sinθ=t2\sin \theta=t2sinθ=t
⇒2cosθ=dtdθ\Rightarrow 2\cos \theta=\dfrac{dt}{d\theta}⇒2cosθ=dθdt
When θ=0,t=0;θ=π/2,t=2\theta=0,t=0; \theta=\pi/2,t=2θ=0,t=0;θ=π/2,t=2
So, I=12∫02f(t)dtI=\frac12\int_0^{2}f(t)dtI=21∫02f(t)dt
=12×6=\frac12\times 6=21×6 [Using (i)]
=3=3=3
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