∫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!!!
+33615 Yes, 1 is the correct answer:
Are you sure you have the limits written correctly in your second integral?
+26367 integrating w substitution
\(\large{\int \limits_{0}^{2} \dfrac{x}{ \sqrt{1+2x^2} }dx}\)
\(\begin{array}{|rcll|} \hline && \mathbf{\large{\int \limits_{x=0}^{x=2} \dfrac{x}{ \sqrt{1+2x^2} }dx}} \\ && \quad \boxed{ \text{substitute: } \\ u = 1+2x^2,\ du=4x\ dx,\\ x=0 \Rightarrow u=1+2\cdot 0^2=1,\\ x=2\Rightarrow u=1+2\cdot 2^2=9 } \\ &=& \int \limits_{u=1}^{u=9} \dfrac{x}{ \sqrt{u} }\dfrac{du}{4x} \\\\ &=& \dfrac{1}{4} \int \limits_{u=1}^{u=9} \dfrac{1}{ \sqrt{u} }du \\\\ &=& \dfrac{1}{4} \int \limits_{u=1}^{u=9} u^{-\frac{1}{2}} du \\\\ &=& \dfrac{1}{4} \left[ \dfrac{u^{\frac{1}{2}}}{\frac{1}{2}} \right]_{u=1}^{u=9} \\\\ &=& \dfrac{1}{4}\cdot \dfrac{2}{1} \left[ u^{\frac{1}{2}} \right]_{u=1}^{u=9} \\\\ &=& \dfrac{1}{2} \left( 9^{\frac{1}{2}}-1^{\frac{1}{2}} \right) \\\\ &=& \dfrac{1}{2} ( \sqrt{9} -1 ) \\\\ &=& \dfrac{1}{2} ( 3 -1 ) \\\\ &=& \dfrac{2}{2} \\\\ &\mathbf{=}& \mathbf{1} \\ \hline \end{array}\)
