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Need help solving this seemingly easy integral
- Thread starter jmj5148
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D
Deleted member 4993
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Hint:jmj5148 said:Integral of dx/(2+.25x^2) from x=0 to 1
Using maple i get an answer of .4806....but everytime i work it out by hand i get half of this answer
...please tell me what im doing wrong, step by step explanation would be nice
Substitute
?(0.125) * x = tan(?)
Show your work step-by-step - for us to be able to tell you what you are doing wrong.
D
Deleted member 4993
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jmj5148 said:in my calc book, the rule for this integral would give you (1/a)arctan(u/a), which would be (1/sqrt(2))arctan(.5x/sqrt(2))........maple and wolfgram gives a number of 1.4141 in front of the arctan, which is just sqrt(2)...
What is your u? What is your a?
D
Deleted member 4993
Guest
jmj5148 said:Integral of dx/(2+.25x^2) from x=0 to 1
Using maple i get an answer of .4806....but everytime i work it out by hand i get half of this answer...please tell me what im doing wrong, step by step explanation would be nice
\(\displaystyle \int_0^1\frac{dx}{2+0.25x^2}\)
\(\displaystyle =\ \frac{1}{2}\int_0^1\frac{dx}{1+0.125x^2}\)
substitute:
?(0.125)*x = tan(?)
and continue....
jmj5148 said:in my calc book, the rule for this integral would give you (1/a)arctan(u/a), which would be (1/sqrt(2))arctan(.5x/sqrt(2))........maple and wolfgram gives a number of 1.4141 in front of the arctan, which is just sqrt(2)...
is (1/a)arctan(u/a) the right way to evalutate this integral with the integral in the form of du/(a^2 + u^2)......cause it doesnt give the right answer but my book and websites give that expression to solve this
BigGlenntheHeavy
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- Mar 8, 2009
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\(\displaystyle Some \ of \ the \ seemingly \ "easy" \ integrals \ aren't \ that \ easy, \ to \ wit:\)
\(\displaystyle \int_{0}^{1}\frac{dx}{2+.25x^2} \ = \ 4 \int_{0}^{1}\frac{dx}{x^2+8}\)
\(\displaystyle Now, \ let \ x \ = \ \sqrt8tan(u), \ dx \ = \ \sqrt8sec^2(u)du\)
\(\displaystyle Hence, \ we \ have \ 4\sqrt8\int_{0}^{arctan(\sqrt2/4)}\frac{sec^2(u)du}{8tan^2(u)+8} \ = \ \sqrt2\int_{0}^{arctan(\sqrt2/4)}\frac{sec^2(u)du}{sec^2(u)}\)
\(\displaystyle = \ \sqrt2 u\bigg]_{0}^{arctan(\sqrt2/4)} \ \ \dot= \ .480601966345\)
\(\displaystyle \int_{0}^{1}\frac{dx}{2+.25x^2} \ = \ 4 \int_{0}^{1}\frac{dx}{x^2+8}\)
\(\displaystyle Now, \ let \ x \ = \ \sqrt8tan(u), \ dx \ = \ \sqrt8sec^2(u)du\)
\(\displaystyle Hence, \ we \ have \ 4\sqrt8\int_{0}^{arctan(\sqrt2/4)}\frac{sec^2(u)du}{8tan^2(u)+8} \ = \ \sqrt2\int_{0}^{arctan(\sqrt2/4)}\frac{sec^2(u)du}{sec^2(u)}\)
\(\displaystyle = \ \sqrt2 u\bigg]_{0}^{arctan(\sqrt2/4)} \ \ \dot= \ .480601966345\)