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Calculus/Integration problem

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Expert: - 10/9/2010


Question
Hello.
Please tell me is the following integration correct and can you explain how is it possible (please tell the basic formula). I saw it written in a book and mulled over it a lot to know how is this integration performed but all in vain.  

∫ (ds / SQRT(s2 - 0.01)) = ln ( SQRT(s2-0.01) + s)

Thanking you in advance.

Answer
No, the is not correct since the derivative of ln(f(s)) = f'(s)/f(s).

Now the ∫(ds/SQRT(s2-0.01)) should really be written as
∫(1/SQRT(s2 - 0.01)) ds, since what comes after the ds is not actually in the integral.

To do this, take a triangle with hypotenuse s and far side 0.1.  This makes the s^2 – 0.01 = s^2 – 0.1^2 be the difference of squares of those two values, so SQRT(s^2-0.01) is the length of the near side.

The equation to use would be s/0.1 = csc(A), where A is the angle, which is 10s = csc(A).  Taking the derivative of both sides gives
10 ds = -csc(A)ctn(A)dA.  This changes to ds = [-csc(A)ctn(A)/10] dA.

Note that 0.1/SQRT(s2 - 0.01) = far side / near side = tan(A).

Multiplying both sides of the last equation by 10 gives
1/SQRT(s2 - 0.01) = 10* tan(A).

We cab now make the substitution back in the integeral.
Then ∫(1/SQRT(s2 - 0.01)) ds becomes
∫ -10*tan(A)* csc(A)ctn(A)/10] dA.

Since tan(A)ctn(A) = 1 and 10/10 = 1, this becomes ∫ -csc(A) dA.
From the book, I believe this to be -ln(csc(A) + ctn(A)).

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