Trigonometric substitutions
- Thread starter flakine
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Trig? Maybe later. Try rewriting it, first. Note that everything is nice and smooth and continuous and positive and finite on the interval you have been given.
\(\displaystyle \L\,\int_{2\sqrt{2}}^{4}{\frac{t}{t^{4}\sqrt{t^{2}-4}}}\,dt\)
Perhaps, after a simpler substitution, maybe \(\displaystyle u^{2} = t^{2} - 4\), it will all be more plain?
\(\displaystyle \L\,\int_{2\sqrt{2}}^{4}{\frac{t}{t^{4}\sqrt{t^{2}-4}}}\,dt\)
Perhaps, after a simpler substitution, maybe \(\displaystyle u^{2} = t^{2} - 4\), it will all be more plain?
Hello, flakine!
Let \(\displaystyle t\,=\,2\sec\theta\;\;\Rightarrow\;\;dt\,=\,2\sec\theta\tan\theta\,d\theta\)
Substitute: \(\displaystyle \L\,\int\frac{2\sec\theta\tan\theta\,d\theta}{(2\sec\theta)^3(2\tan\theta)} \;=\;\frac{1}{8}\int\frac{d\theta}{\sec^2\theta}\;=\;\frac{1}{8}\int\)\(\displaystyle \cos^2\theta\,d\theta \;= \;\frac{1}{16}\L\int\)\(\displaystyle \left(1\,+\,\cos2\theta\right)\,d\theta\)
Can you finish it now?
Let \(\displaystyle t\,=\,2\sec\theta\;\;\Rightarrow\;\;dt\,=\,2\sec\theta\tan\theta\,d\theta\)
Substitute: \(\displaystyle \L\,\int\frac{2\sec\theta\tan\theta\,d\theta}{(2\sec\theta)^3(2\tan\theta)} \;=\;\frac{1}{8}\int\frac{d\theta}{\sec^2\theta}\;=\;\frac{1}{8}\int\)\(\displaystyle \cos^2\theta\,d\theta \;= \;\frac{1}{16}\L\int\)\(\displaystyle \left(1\,+\,\cos2\theta\right)\,d\theta\)
Can you finish it now?
- Joined
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Really strong hint.
\(\displaystyle \L\,u^{2} = t^{2} - 4\)
\(\displaystyle \L\,u^{2} + 4 = t^{2}\)
\(\displaystyle \L\,(u^{2} + 4)^{2} = (t^{2})^{2} = t^{4}\)
You can get to the trig substitution in a moment. This will just get you there.
\(\displaystyle \L\,u^{2} = t^{2} - 4\)
\(\displaystyle \L\,u^{2} + 4 = t^{2}\)
\(\displaystyle \L\,(u^{2} + 4)^{2} = (t^{2})^{2} = t^{4}\)
You can get to the trig substitution in a moment. This will just get you there.