MHB How to calculate this type of integral

[zhaojx84]

Could anyone can tell me how to calculate this type of intergretion. Thanks very much
$$\int\frac{{y}^{3}}{(196 - {y}^{2})\times \sqrt{196 - {y}^{2} - {a + y}^{2}}}$$
View attachment 7724

 

[Euge]

Re: How to calculate this type of intergretion

Your integral does not match the integral in the picture. Which one are you trying to evaluate?

 

[Prove It]

Could anyone can tell me how to calculate this type of intergretion. Thanks very much
$$\int\frac{{y}^{3}}{(196 - {y}^{2})\times \sqrt{196 - {y}^{2} - {a + y}^{2}}}$$

I'll assume that the picture is the actual integral you are trying to solve...

$\displaystyle \begin{align*} \int{ \frac{y^3}{\left( 196 - y^2 \right) \,\sqrt{-\left( a + y \right) ^2 - y^2 + 196}} \,\mathrm{d}y } &= -\frac{1}{2} \int{ \frac{y^2}{\left( 196 - y^2 \right) \,\sqrt{ -\left( a + y \right) ^2 - y^2 + 196 }} \,\left( -2\,y \right) \,\mathrm{d}y } \end{align*}$

Let $\displaystyle \begin{align*} u = 196 - y^2 \implies \mathrm{d}u = -2\,y\,\mathrm{d}y \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} -\frac{1}{2} \int{\frac{y^2}{\left( 196 - y^2 \right) \,\sqrt{-\left( a + y \right) ^2 - y^2 + 196}}\,\left( -2\,y \right) \,\mathrm{d}y} &= -\frac{1}{2} \int{ \frac{196 - u}{u\,\sqrt{ \left( a + u - 196 \right) ^2 - u }} \,\mathrm{d}u } \end{align*}$

Does this seem a bit more manageable?

 

[zhaojx84]

Re: How to calculate this type of intergretion

Your integral does not match the integral in the picture. Which one are you trying to evaluate?

The integration in the figure is what I want to calculate.
Thanks very much for your help.

 

[zhaojx84]

I'll assume that the picture is the actual integral you are trying to solve...

$\displaystyle \begin{align*} \int{ \frac{y^3}{\left( 196 - y^2 \right) \,\sqrt{-\left( a + y \right) ^2 - y^2 + 196}} \,\mathrm{d}y } &= -\frac{1}{2} \int{ \frac{y^2}{\left( 196 - y^2 \right) \,\sqrt{ -\left( a + y \right) ^2 - y^2 + 196 }} \,\left( -2\,y \right) \,\mathrm{d}y } \end{align*}$

Let $\displaystyle \begin{align*} u = 196 - y^2 \implies \mathrm{d}u = -2\,y\,\mathrm{d}y \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} -\frac{1}{2} \int{\frac{y^2}{\left( 196 - y^2 \right) \,\sqrt{-\left( a + y \right) ^2 - y^2 + 196}}\,\left( -2\,y \right) \,\mathrm{d}y} &= -\frac{1}{2} \int{ \frac{196 - u}{u\,\sqrt{ \left( a + u - 196 \right) ^2 - u }} \,\mathrm{d}u } \end{align*}$

Does this seem a bit more manageable?

Thanks for your help.
However, the substitution in the sqrt root is not correct.

 

[Prove It]

Thanks for your help.
However, the substitution in the sqrt root is not correct.

Yes I shouldn't try to help while tired haha. I'm sure you can fix it though :)

 

[zhaojx84]

Yes I shouldn't try to help while tired haha. I'm sure you can fix it though :)

Could you help me later? I cannot figure it out. Thanks very much!