- #1
ritwik06
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Homework Statement
PLease HELP ME INTEGRATE THIS>
[tex]\int\frac{dh}{\sqrt{h^{2}-k^{2}}}[/tex]
k is constant
ritwik06 said:I am redefining the question:
[tex]\int\frac{dx}{\sqrt{x^{2}-a^{2}}}[/tex]
taking x^2 common
[tex]\int\frac{dx}{x\sqrt{1-\frac{a^{2}}{x^{2}}}}[/tex]
let [tex]1-\frac{a^{2}}{x^{2}}=t[/tex]
[tex]\frac{1}{2}\int\frac{dt}{(1-t)(\sqrt{t})}[/tex]
let[tex]\sqrt{t}=j[/tex]
[tex]\int\frac{dj}{1-j^{2}}[/tex]
how shall i proceed
Pretty much every step is wrong. You seem to be consistly forgetting to replace the "dx" or "dt" term.ritwik06 said:I am redefining the question:
[tex]\int\frac{dx}{\sqrt{x^{2}-a^{2}}}[/tex]
taking x^2 common
[tex]\int\frac{dx}{x\sqrt{1-\frac{a^{2}}{x^{2}}}}[/tex]
let [tex]1-\frac{a^{2}}{x^{2}}=t[/tex]
[tex]\frac{1}{2}\int\frac{dt}{(1-t)(\sqrt{t})}[/tex]
let[tex]\sqrt{t}=j[/tex]
[tex]\int\frac{dj}{1-j^{2}}[/tex]
how shall i proceed
Integration with k as constant is a mathematical process that involves finding the antiderivative or integral of a function where k is treated as a constant. This means that k does not change and can be treated as a fixed value throughout the integration process.
K is treated as a constant in integration because it simplifies the integration process. It allows us to treat k as a single value rather than a changing variable, making it easier to solve the integral and obtain a more precise result.
Integration with k as constant is different from other integration methods because it involves treating k as a constant rather than a variable. Other integration methods may involve treating other constants or variables as changing values, which can make the integration process more complex.
Using k as a constant in integration can help simplify the process and make it easier to solve. It also allows for more accurate and precise results, as treating k as a constant eliminates any errors that may arise from treating it as a variable.
Yes, k can be treated as a variable in integration. However, this may make the integration process more complex and may result in less accurate solutions. It is often beneficial to treat k as a constant in integration to simplify the process and obtain more precise results.