Why do we set $x+1=\frac{1}{t}$ When we compute $I= \int_{{1}/{2}}^{0} \frac{dx}{(x+1)\sqrt{(3-x)(x+1)}}.$

Asked Mar 25 '15 at 16:14

Active Jan 28 '19 at 06:26

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When we compute $I=\displaystyle \int\limits_{\frac{1}{2}}^{0} \dfrac{dx}{(x+1)\sqrt{(3-x)(x+1)}}.$
We set $x+1=\dfrac{1}{t}$ and we have $\displaystyle I=\int\limits_1^2\dfrac{dt}{\sqrt{4t-1}}$.

I have a question : "Why do we set $x+1=\dfrac{1}{t}$?"

Riemann integral.

edited Jan 28 '19 at 06:26

hmakholm left over Monica

286,031

asked Mar 25 '15 at 16:14

Hung nguyen

You don't have to. There are many other transformations that also work. – Nathanson Mar 25 '15 at 16:15
@Nathanson@ can you show for me a different transformation? – Hung nguyen Mar 26 '15 at 02:31
The third one of these also works. – Nathanson Mar 26 '15 at 03:27

Why do we set $x+1=\frac{1}{t}$ When we compute $I= \int_{{1}/{2}}^{0} \frac{dx}{(x+1)\sqrt{(3-x)(x+1)}}.$

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