- #1
Hertz
- 180
- 8
Homework Statement
[itex]\int \frac{dx}{(1+4x^2)^{3/2}}[/itex]
The Attempt at a Solution
[itex]\int \frac{dx}{(1+4x^2)^{3/2}}[/itex]
Let [itex]x = \frac{1}{4}tan(u), dx = \frac{1}{4}sec^2(u)du[/itex]
[itex]\frac{1}{4}\int{\frac{sec^2(u)du}{(sec^2(u))^{3/2}}}[/itex]
[itex]\frac{1}{4}\int{\frac{1}{sec(u)}}du[/itex]
[itex]\frac{1}{4}\int{cos(u)}du[/itex]
[itex]\frac{1}{4}sin(u) + C[/itex]Drawing a triangle with angle u:[itex]tan(u) = 4x[/itex]
Therefore:
Opposite = [itex]4x[/itex]
Adjacent = [itex]1[/itex]
Hypotenuse = [itex]\sqrt{1^2 + (4x)^2} = \sqrt{1 + 16x^2}[/itex]
[itex]sin(u) = \frac{4x}{\sqrt{1 + 16x^2}}[/itex][itex]\frac{1}{4}sin(u) + C[/itex]
[itex]\frac{1}{4}(\frac{4x}{\sqrt{1 + 16x^2}}) + C[/itex]
[itex]\frac{x}{\sqrt{1 + 16x^2}} + C[/itex]
I personally can't see what I did wrong here, but this is not the correct answer :( Both my math book and Mathematica say that the correct answer should actually be
[itex]\frac{x}{\sqrt{1 + 4x^2}} + C[/itex]
Any help is greatly appreciated :)
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