The integral of ( \cos x \sin x ) can be computed using a trigonometric identity. We use the identity ( \sin(2x) = 2 \sin x \cos x ), which allows us to rewrite the integral as:
[ \int \cos x \sin x , dx = \frac{1}{2} \int \sin(2x) , dx. ]
Integrating ( \sin(2x) ) gives:
[ \frac{-1}{2} \cos(2x) + C, ]
thus the final result is:
[ \int \cos x \sin x , dx = \frac{-1}{4} \cos(2x) + C. ]
2
In reimann stieltjes integral if we assume a(x) = x then it becomes reimann integral so we can say R-S integral is generalized form of reimann integral.
Political parties are an integral part of democracy. Religion is an integral part of Saudi Arabia. Greed is an integral part of human psychology.
Particular integral is finding what the integral is for example the integral of 2x is x^2 + C. Finding the particular solution would be finding what C equals from the particular integral.
It's not. It depends on the method you use for summation whether summation > integral or integral > summation.
2
Without specifying the limits of integration, the integral will always include an arbitrary constant, and you'll never get a numerical value for it. So the statement is untrue on its face, hence unprovable. We'll have a look at the trig later.
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You know that cosπ/4 and sinπ/4 both equal 1/root2, so multiplying them gets 1/2.
8
Integral in Tagalog: mahalaga
In reimann stieltjes integral if we assume a(x) = x then it becomes reimann integral so we can say R-S integral is generalized form of reimann integral.
Elections are integral to democracies.
Do you mean the Convolution Integral?
What is a integral
non integral is type of numbers behaviour: i can say that set of numbers without any "holes inside" are integral and set of numbers with "holes inside are non integral. example : integral group "1..100" non integral group "1,4,8,67"
Integral of cos^2x=(1/2)(cosxsinx+x)+CHere is why:Here is one method: use integration by parts and let u=cosx and dv=cosxdxdu=-sinx v=sinxInt(udv)=uv-Int(vdu) so uv=cosx(sinx) and vdu=sinx(-sinx)so we have:Int(cos^2(x)=(cosx)(sinx)+Int(sin^2x)(the (-) became + because of the -sinx, so we add Int(vdu))Now it looks not better because we have sin^2x instead of cos^2x,but sin^2x=1-cos^2x since sin^2x+cos^2x=1So we haveInt(cos^2x)=cosxsinx+Int(1-cos^2x)=cosxsinx+Int(1)-Int(cos^2x)So now add the -Int(cos^2x) on the RHS to the one on the LHS2Int(cos^2x)=cosxsinx+xso Int(cos^2x)=1/2[cosxsinsx+x] and now add the constant!final answerIntegral of cos^2x=(1/2)(cosx sinx + x)+C = x/2 + (1/4)sin 2x + C(because sin x cos x = (1/2)sin 2x)Another method is:Use the half-angle identity, (cos x)^2 = (1/2)(1 + cos 2x). So we have:Int[(cos x)^2 dx] = Int[(1/2)(1 + cos 2x)] dx = (1/2)[[Int(1 dx)] + [Int(cos 2x dx)]]= (1/2)[x + (1/2)sin 2x] + C= x/2 +(1/4)sin 2x + C