I want you to look at question (b) in the picture provided and tell me where (1/3) outside the integral came from. Is there factoring involved
I am to use the formula below.
Let INT = integral
INT (du)/(a^2+u^2) = (1/a)*arctan(u/a) + C
I want you to look at question (b) in the picture provided and tell me where (1/3) outside the integral came from. Is there factoring involved
I am to use the formula below.
Let INT = integral
INT (du)/(a^2+u^2) = (1/a)*arctan(u/a) + C
It's because when the substitution $\displaystyle \begin{align*} u = 3x \end{align*}$ is applied, the differential is $\displaystyle \begin{align*} du = 3\,dx \end{align*}$. To get the extra factor of 3 on the inside, a factor of 1/3 has to be multiplied outside.
$\displaystyle{\int}\dfrac{dx}{2+9x^2}$
and you want to use the template
$\displaystyle{\int}\dfrac{dx}{x^2+a^2}=\dfrac{1}{ a}\arctan\left(\frac{x}{a}\right)+C$
$\large \begin{align*}
&\displaystyle{\int}\dfrac{dx}{2+9x^2}= \\ \\
&\frac{1}{9}\displaystyle{\int}\dfrac{dx}{\frac{2} {9}+x^2} =\\ \\
&\frac{1}{9}\displaystyle{\int}\dfrac{dx}{ \left(\sqrt{ \frac{2}{9}}\right)^2+x^2} \\ \\
&\mbox{we see that }a=\sqrt{\frac{2}{9}} \\ \\
&\frac{1}{9}\displaystyle{\int}\dfrac{dx}{ \left(\sqrt{ \frac{2}{9}}\right)^2+x^2}= \\ \\
&\frac{1}{9} \sqrt{\dfrac{9}{2}}\arctan\left(\sqrt{ \frac{9}{2}}x\right)+C = \\ \\
&\frac{1}{3\sqrt{2}} \arctan\left(\frac{3x}{\sqrt{2}}\right)+C
\end{align*}$