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Compute the indefinite integral ∫cos(x)/(1+sin(x)+cos(x)).
= arctan(tan(x/2)) - (1/2) ln(tan^2(x/2) + 1) + C.
Then compute the boundaries.
= 1/4 * (π - ln(4)) - 0
= 1/4 * (π - ln(4))
= arctan(tan(x/2)) - (1/2) ln(tan^2(x/2) + 1) + C.
Then compute the boundaries.
= 1/4 * (π - ln(4)) - 0
= 1/4 * (π - ln(4))
kvnmurty:
the second line, integrand is correct. But is it in a some what complex form. It could be simplified further.
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integral of
![f(x)=\frac{cos x}{1+Cosx+Sinx}\\\\=\frac{cos^2\frac{x}{2}-sin^2\frac{x}{2}}{2Cos^2\frac{x}{2}+2Sin\frac{x}{2}Cos\frac{x}{2}}\\\\=\frac{(cos\frac{x}{2}+sin\frac{x}{2})(cos\frac{x}{2}-sin\frac{x}{2})}{(cos\frac{x}{2}+sin\frac{x}{2})(2cos\frac{x}{2})}\\\\=\frac{1}{2}-\frac{sin\frac{x}{2}}{cos\frac{x}{2}}*\frac{1}{2}\\\\=\ \textgreater \ \ \int\limits^{\frac{\pi}{2}}_0 {f(x)} \, dx=[\frac{x}{2}+Ln(Cos\frac{x}{2})]_0^\frac{\pi}{2}\\\\\frac{\pi}{4}-\frac{1}{2}Ln_e 2.](https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7Bcos+x%7D%7B1%2BCosx%2BSinx%7D%5C%5C%5C%5C%3D%5Cfrac%7Bcos%5E2%5Cfrac%7Bx%7D%7B2%7D-sin%5E2%5Cfrac%7Bx%7D%7B2%7D%7D%7B2Cos%5E2%5Cfrac%7Bx%7D%7B2%7D%2B2Sin%5Cfrac%7Bx%7D%7B2%7DCos%5Cfrac%7Bx%7D%7B2%7D%7D%5C%5C%5C%5C%3D%5Cfrac%7B%28cos%5Cfrac%7Bx%7D%7B2%7D%2Bsin%5Cfrac%7Bx%7D%7B2%7D%29%28cos%5Cfrac%7Bx%7D%7B2%7D-sin%5Cfrac%7Bx%7D%7B2%7D%29%7D%7B%28cos%5Cfrac%7Bx%7D%7B2%7D%2Bsin%5Cfrac%7Bx%7D%7B2%7D%29%282cos%5Cfrac%7Bx%7D%7B2%7D%29%7D%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D-%5Cfrac%7Bsin%5Cfrac%7Bx%7D%7B2%7D%7D%7Bcos%5Cfrac%7Bx%7D%7B2%7D%7D%2A%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%3D%5C+%5Ctextgreater+%5C+%5C+%5Cint%5Climits%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D_0+%7Bf%28x%29%7D+%5C%2C+dx%3D%5B%5Cfrac%7Bx%7D%7B2%7D%2BLn%28Cos%5Cfrac%7Bx%7D%7B2%7D%29%5D_0%5E%5Cfrac%7B%5Cpi%7D%7B2%7D%5C%5C%5C%5C%5Cfrac%7B%5Cpi%7D%7B4%7D-%5Cfrac%7B1%7D%7B2%7DLn_e+2. "f(x)=\frac{cos x}{1+Cosx+Sinx}\\\\=\frac{cos^2\frac{x}{2}-sin^2\frac{x}{2}}{2Cos^2\frac{x}{2}+2Sin\frac{x}{2}Cos\frac{x}{2}}\\\\=\frac{(cos\frac{x}{2}+sin\frac{x}{2})(cos\frac{x}{2}-sin\frac{x}{2})}{(cos\frac{x}{2}+sin\frac{x}{2})(2cos\frac{x}{2})}\\\\=\frac{1}{2}-\frac{sin\frac{x}{2}}{cos\frac{x}{2}}*\frac{1}{2}\\\\=\ \textgreater \ \ \int\limits^{\frac{\pi}{2}}_0 {f(x)} \, dx=[\frac{x}{2}+Ln(Cos\frac{x}{2})]_0^\frac{\pi}{2}\\\\\frac{\pi}{4}-\frac{1}{2}Ln_e 2.")
You could also express cos x and sin x in terms of tan x/2 and then say u = tan x/2. Express integrand in terms of u. and integrate that.
You could also express cos x and sin x in terms of tan x/2 and then say u = tan x/2. Express integrand in terms of u. and integrate that.
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