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About Abe Mantell
Expertise
Hello, I am a college professor of mathematics and regularly teach all levels from elementary mathematics through differential equations, and would be happy to assist anyone with such questions!

Experience
Over 15 years teaching at the college level.

Organizations
NCTM, NYSMATYC, AMATYC, MAA, NYSUT, AFT.

Education/Credentials
B.S. in Mathematics from Rensselaer Polytechnic Institute
M.S. (and A.B.D.) in Applied Mathematics from SUNY @ Stony Brook

You are here: Experts > Teens > Homework/Study Tips > Calculus > Calculus

Topic: Calculus

Expert: Abe Mantell
Date: 8/29/2005
Subject: Calculus

Question
Dear Abe,

Thanks for your help. I have used the rule:
Integral(dx / x^2 - a^2) = 1/2a ln (absolute vale of)(x-a)/(x-b) + c
to get

Integral (sin(x)/(1+sin^2(x)) dx)
= - ((ln(cosx - sqrt2) - ln(cosx + sqrt2)) / (2 sqrt2) + c
Hopefully this makes sense to you.

If the limits of Integral (sin(x)/(1+sin^2(x)) dx) are between pi/2 and 0,
is it correct to use u=0 when x=pi/2 & u=1 when x=0 to get -1.4351?

Thanks again,
Allison.
-------------------------
Followup To
Question -
How is the integral of sin x / 1+sin^2 x found?
Answer -
Hello Allison,

I gathe you mean:
Integral(sin(x)/(1+sin^2(x)) dx)...yes?

If so, then let u=cos(x), so that du=-sin(x)dx
and sin^2(x)=1-u^2...so the integral becomes:
Integral(du/(u^2-2))...can you finish it from here?

Let me know if you need more help! :-)

Abe

Answer
Hello again Allison,

Should the limits of integration go from
x=0 TO x=pi/2 (rather than the reverse)?

If so, then I get the integral is:
-sqrt(2)ln(3-2sqrt(2))/4
which is about 0.623225

If the limits of integration are reversed, as you
have them, then the answer is just the negative
of what I have above...OK?

TTYL, Abe

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