# Integration by Substitution, Converting dx to du

Q: Using the substitution of $$u = \dfrac{e^{2x}}{5}$$, write this integrand as a function of u: $$\int \frac{e^{2x}}{\sqrt{25-e^{4x}}}dx$$

I'm stuck at substituting u into the denominator. The best I can get is: $$\int \frac{5u}{\sqrt{25-{ \frac {5}{2} (\frac{2}{5}e^{4x}})}}du$$

I've tried making it $$\int \frac{5u}{\sqrt{25-{ \frac {5}{2} (\frac{2}{5}e^{2x}})^2}}du$$ but i guess it's wrong too as the items in the bracket will also be squared which is wrong.

Can anyone help me with this? Thank you!

• $e^{4x}=(e^{2x})^{2}=25u^{2}$. Also $dx=\frac 5 {2u} du$. – Kabo Murphy May 21 at 6:09
• @KaviRamaMurthy Thank you!!! – user672518 May 21 at 6:12
• @KaviRamaMurthy Why are you answering in a comment? – Arthur May 21 at 6:14

When you say $$5u=e^{2x}$$ then $$5du=2e^{2x}dx$$ or $$\dfrac52du=e^{2x}dx$$ so $$\int \frac{e^{2x}dx}{\sqrt{25-e^{4x}}} = \int \dfrac{\dfrac52du}{\sqrt{25-25u^2}}=\dfrac12\int\dfrac{du}{\sqrt{1-u^2}}$$
Let $$e^{2x}=5\sin\theta$$.