perplexus.info :: Calculus : Equate The Integrals, Get Constant

Equate The Integrals, Get Constant (Posted on 2007-10-24)

Determine the value of a constant C such that:

∫₀^{^pi/3}(sin y/cos²y)dy = ∫₀^{^C}(√(z+C) - √z)^-1dz

Note: The range of the first integral reads 'pi/3'.

Submitted by K Sengupta

Rating: 2.5000 (2 votes)

Solution: (Hide)

C = 9/32

EXPLANATION:

Substituting t = sec y, we obtain dt = sec y .tan y.dy

Thus, the integral on the right is:
integral(dt), t = 1 to 2
= 2-1 = 1

Also, the integral on the left:
= integral (V(z+C) – VC)^-1.dz, z = 0 to C
= (1/C)* integral (V(z+C) + VC)^-1.dz, z = 0 to C
= 4*(V2/3)*VC

Thus, 32*C/9 = 1, giving C = 9/32, so that:

The required value of C is 9/32.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Sorry is the same result......!	Chesca Ciprian	2007-10-25 15:36:44
Hmmm....	Chesca Ciprian	2007-10-25 15:34:10
full solution	Daniel	2007-10-24 18:18:26

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