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FilipVz
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Can somebody explain how to solve integral from the picture above?( solution is in the second line)
Can somebody explain how to solve integral from the picture above?( solution is in the second line)
MarkFL said:Hint: Try the substitution:
\(\displaystyle \cot(\theta)=\frac{\sqrt{1-k^2}}{k}\cos(u)\)
Hence:
\(\displaystyle \csc^2(\theta)\,d\theta=\frac{\sqrt{1-k^2}}{k}\sin(u)\,du\)
And the result will follow. :D
FilipVz said:Thank you,
i solved it, using substitution:
\(\displaystyle u= (k*ctgθ)/(1-k^2 )\)
But, my result is:
\(\displaystyle ϕ=-arcsin((k∙ctgθ)/√(1-k^2 ))+c_2\)
instead of
\(\displaystyle ϕ=arccos((k∙ctgθ)/√(1-k^2 ))+c_2\)
Is this correct?
To solve an integral using a picture, you need to first determine the bounds of integration and the function being integrated. Then, you can graph the function and the area under the curve. Next, you can use geometric shapes to estimate the area and break it down into smaller, easier to solve integrals. Finally, you can use the fundamental theorem of calculus to find the exact value of the integral.
The fundamental theorem of calculus states that the definite integral of a function f(x) can be evaluated by finding the antiderivative of f(x) and evaluating it at the upper and lower bounds of integration. In other words, it relates differentiation and integration by showing that integration is the inverse operation of differentiation.
No, using a picture to solve integrals is most effective for finding the area under a curve for continuous functions. It may not work for more complex integrals involving trigonometric, logarithmic, or exponential functions.
Using a picture to solve integrals is not always accurate and is more of an estimation method. The accuracy depends on the precision of the graph and the geometric shapes used to estimate the area. It is recommended to use other methods, such as the fundamental theorem of calculus, for more precise results.
Yes, there are some limitations to using a picture to solve integrals. It may not work for all types of functions and can be time-consuming and error-prone. Also, it may not be feasible to use this method for more complex integrals with multiple variables or in higher dimensions.