∫^{20 }(x/sqrt(1+2x^2))dx

I got 1 as my final answer....but I'm not sure if thats correct. Even if it is right, would someone be able to go through this problem step by step?

also if anyone could help me with this following problem:

∫^{3π^2 π/2 }cosx/sin^2(x)dx

thank you!!!

Guest Mar 25, 2019

edited by
Guest
Mar 25, 2019

edited by Guest Mar 25, 2019

edited by Guest Mar 25, 2019

edited by Guest Mar 25, 2019

edited by Guest Mar 25, 2019

#1**+3 **

Yes, 1 is the correct answer:

Are you sure you have the limits written correctly in your second integral?

Alan Mar 25, 2019

#2**+2 **

**integrating w substitution**

\(\large{\int \limits_{0}^{2} \dfrac{x}{ \sqrt{1+2x^2} }dx}\)

\(\begin{array}{|rcll|} \hline && \mathbf{\large{\int \limits_{x=0}^{x=2} \dfrac{x}{ \sqrt{1+2x^2} }dx}} \\ && \quad \boxed{ \text{substitute: } \\ u = 1+2x^2,\ du=4x\ dx,\\ x=0 \Rightarrow u=1+2\cdot 0^2=1,\\ x=2\Rightarrow u=1+2\cdot 2^2=9 } \\ &=& \int \limits_{u=1}^{u=9} \dfrac{x}{ \sqrt{u} }\dfrac{du}{4x} \\\\ &=& \dfrac{1}{4} \int \limits_{u=1}^{u=9} \dfrac{1}{ \sqrt{u} }du \\\\ &=& \dfrac{1}{4} \int \limits_{u=1}^{u=9} u^{-\frac{1}{2}} du \\\\ &=& \dfrac{1}{4} \left[ \dfrac{u^{\frac{1}{2}}}{\frac{1}{2}} \right]_{u=1}^{u=9} \\\\ &=& \dfrac{1}{4}\cdot \dfrac{2}{1} \left[ u^{\frac{1}{2}} \right]_{u=1}^{u=9} \\\\ &=& \dfrac{1}{2} \left( 9^{\frac{1}{2}}-1^{\frac{1}{2}} \right) \\\\ &=& \dfrac{1}{2} ( \sqrt{9} -1 ) \\\\ &=& \dfrac{1}{2} ( 3 -1 ) \\\\ &=& \dfrac{2}{2} \\\\ &\mathbf{=}& \mathbf{1} \\ \hline \end{array}\)

heureka Mar 25, 2019