definite integral, from 0 to 1, of x^3 / (4 + x^2)^(1/2)
- Thread starter clw89
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Hello, clw89!
I would try Trig Substitution . . .
\(\displaystyle \text{Let: }\,x \:=\:\2tan\theta\quad\Rightarrow\quad dx \:=\:2\sec^2\!\theta\,d\theta\)
. . \(\displaystyle \text{and: }\:\sqrt{4+x^2} \:=\:2\sec\theta\)
\(\displaystyle \text{Substitute: }\;\int\frac{8\tan^3\!\theta}{2\sec\theta}\,(2\sec^2\!\theta\,d\theta) \;=\;8\int\sec\theta\tan^3\!\theta\,d\theta\)
. . \(\displaystyle =\;8\int\tan^2\!\theta(\sec\theta\tan\theta\,d\theta) \;=\;8\int(\sec^2\!\theta-1)(\sec\theta\tan\theta\,d\theta)\)
. . \(\displaystyle \text{Now let }\,u \:=\:\sec\theta\) . . .
I would try Trig Substitution . . .
\(\displaystyle \text{Let: }\,x \:=\:\2tan\theta\quad\Rightarrow\quad dx \:=\:2\sec^2\!\theta\,d\theta\)
. . \(\displaystyle \text{and: }\:\sqrt{4+x^2} \:=\:2\sec\theta\)
\(\displaystyle \text{Substitute: }\;\int\frac{8\tan^3\!\theta}{2\sec\theta}\,(2\sec^2\!\theta\,d\theta) \;=\;8\int\sec\theta\tan^3\!\theta\,d\theta\)
. . \(\displaystyle =\;8\int\tan^2\!\theta(\sec\theta\tan\theta\,d\theta) \;=\;8\int(\sec^2\!\theta-1)(\sec\theta\tan\theta\,d\theta)\)
. . \(\displaystyle \text{Now let }\,u \:=\:\sec\theta\) . . .
skeeter
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\(\displaystyle \int_0^1 \frac{x^3}{\sqrt{4+x^2}} dx\)
\(\displaystyle u = 4 + x^2\)
\(\displaystyle x^2 = u - 4\)
\(\displaystyle du = 2x \, dx\)
\(\displaystyle \frac{1}{2} \int_0^1 \frac{x^2 \cdot 2x}{\sqrt{4+x^2}} dx\)
\(\displaystyle \frac{1}{2} \int_4^5 \frac{u-4}{\sqrt{u}} \, du\)
\(\displaystyle \frac{1}{2} \int_4^5 u^{\frac{1}{2}} - 4u^{-\frac{1}{2} \, du\)
\(\displaystyle u = 4 + x^2\)
\(\displaystyle x^2 = u - 4\)
\(\displaystyle du = 2x \, dx\)
\(\displaystyle \frac{1}{2} \int_0^1 \frac{x^2 \cdot 2x}{\sqrt{4+x^2}} dx\)
\(\displaystyle \frac{1}{2} \int_4^5 \frac{u-4}{\sqrt{u}} \, du\)
\(\displaystyle \frac{1}{2} \int_4^5 u^{\frac{1}{2}} - 4u^{-\frac{1}{2} \, du\)