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Leaping antalope
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can anyone help me with this problem...
the integral from 0 to half pi: 1/(1+cosx)dx...
Thanks...
the integral from 0 to half pi: 1/(1+cosx)dx...
Thanks...
Leaping antalope said:so 1/(1+cosx)=(1-cosx)/(1-cos^2x)=(1-cosx)/sin^2x=(1/sin^2x)-(cosx/sin^2x)...
Aha~now i get it
the integration of 1/sin^2x is -cotx, and cosx/sin^2x=cosx*sin^(-2)x, and the integral of this is -1/sinx
so the integration of 1/(1+cosx) is -cotx+1/sinx, but now i have a problem, cot(half pi) does not exist...
help still needed...
Leaping antalope said:The answer is 1. Or maybe we should convert (1/sin^2x)-(cosx/sin^2x) so that we can substitute half pi and 0 in it...not sure...
arildno said:Hint use the relation:
[tex]\cos^{2}(\frac{x}{2})=\frac{1+\cos(x)}{2}[/tex]
marlon said:I suggest you use the advice of arildno. It will be the best way to solve this.
[tex]\int\frac{2}{\cos(x)+1}*dx = \int\frac{1}{\cos^{2}(\frac{x}{2})} * dx[/tex]
and you know that [tex]dx = 2*d(\frac{x}{2})[/tex]
and
[tex]\int\frac{1}{\cos^{2}(x)} = \tan(x)[/tex]
regards
marlon
marlon said:I suggest you use the advice of arildno. It will be the best way to solve this.
[tex]\int\frac{2}{\cos(x)+1}*dx = \int\frac{1}{\cos^{2}(\frac{x}{2})} * dx[/tex]
and you know that [tex]dx = 2*d(\frac{x}{2})[/tex]
and
[tex]\int\frac{1}{\cos^{2}(x)} = \tan(x)[/tex]
regards
marlon
A definite integral is a mathematical concept used in calculus to find the area under a curve between two specific points on the x-axis. In other words, it is a way to calculate the total accumulation of a function over a given interval.
To solve a definite integral, you can use various techniques such as substitution, integration by parts, or trigonometric identities. In this specific problem, we can use the substitution method to simplify the integral.
The substitution method is a technique used in integration to simplify the integral by substituting a variable with a new expression. In this problem, we can use the substitution u = 1 + cosx to simplify the integral.
To apply the substitution method, we first substitute u = 1 + cosx in the integral, which gives us the integral of 1/u du from 0 to pi/2. Then, we can use the power rule for integrals to solve the integral, which gives us ln(1+cosx) evaluated from 0 to pi/2. Finally, we substitute back the original variable x to get the final answer.
The final solution to this definite integral is ln(2) - ln(1/2) which simplifies to ln(4). This means the area under the curve of 1/(1+cosx) from 0 to pi/2 is ln(4) units squared.