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Calculus/integration

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Expert: - 12/29/2010


Question
hello, i was hoping you could explain what the integral of 1/((u^2/4)+1) is and also the formula used. the question was very different however it is just this line that im stuck on! thankyou in advance.

Answer
Make a right triangle with the are side u/2, the near side 1, and the angle be A.
This would make tan(a) = u/2.
Taking the derivative of both sides gives sec^2(A)dA = (1/2)du.
Multiplying both sides by 2 gives 2*sec^2(A)dA = du.

Note that 1/((u^2/4)+1) is the same as cos^2(A).

Thus, we have what we need to change the integral of 1/((u^2/4)+1) du into
the integral of cos^2(A)[2*sec^2(A)]dA.

Since cos(A) = 1/sec(A), the cos^2(A)*sec^2(A) = 1, so all we have is the
integral 2 dA.

This integrates to 2A + C, and A is known to be arctan(u/2).
This gives us a final answer of 2*arctan(u/2) + C.  

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