0
cos(3t) = cos(2t + t) = cos(2t)*cos(t) - sin(2t)*sin(t)
= [cos2(t) - sin2(t)]*cos(t) - 2*cos(t)*sin(t)*sin(t)
= [cos2(t) - sin2(t)]*cos(t) - 2*cos(t)*sin2(t)
then, since sin2(t) = 1 - cos2(t)
= [2*cos2(t) - 1]*cos(t) - 2*cos(t)*[1 - cos2(t)]
= 2*cos3(t) - cos(t) - 2*cos(t) + 2*cos3(t)
= 4*cos3(t) - 3*cos(t)
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Integral of (sin2x)(cos3x)?
Int[sin(2x)*cos(3x)]dx = Int[(2sinx*cosx)*(4cos^3x - 3cosx)]dx= Int[(8sinx*cos^4x - 6sinx*cos^2x)]dx Let cosx = u then du/dx = -sinx So, the integral is Int[-8*u^4 + 6*u^2]du = -8/5*u^5 + 2u^3 + c where c is a constant of integration = -8/5*cos^5x + 2cos^3x + c
What is the derivative of the square root of cos x?
To find the derivatve of the square root of cos x: Use the chain rule; this means multiply the inner derivative by the outer derivative. You can write the question f(x) = (cos x)1/2 This general break-down explains how to find d/dx f(x) note: d/dx basically symbolizes "the derivative of" In general terms: f(x) = x1/2 g(x) = cos x f(g(x)) = (cos x)1/2 outer derivative: d/dx f(z) = (1/2)*x-1/2 = 1/(4cos x)1/2 inner derivative: d/dx g(x) = -sin(x) final answer: d/dx f(g(x)) = -sin(x)/(4*cos x)1/2 note: d/dx means "the derivative of"; so d/dx x = 1 Further explained: Set up the equation to a more general form: (cos x)1/2 To make the inner derivative, look at cos(x) To make the outer derivative, look at x1/2 note: x ~ cos x; so we treat (cos x) simply as x, to create the outer derivative You probably know the necessary derivates: 1. derivative of cos x = -sin x 2. derivative of a1/2 = (1/2)*a-1/2 = 1/(4a)1/2 Multiplying the two we get the answer: -sin(x)/(4cos x)1/2
What is the geometrical interpretation of line integral?
A line integral or a path or curve integral is a function used when we integrate along a curve. The first step is to parametrize the curve if it is not already in parametric form. So let's look at a general curve, c in its parametric form. The letter t will stand for time. x=h(t) and y=g(t) so both x and y are functions of time. (We assume the curve is smooth) We will integrate over an interval [a,b] so a≤t≤b Now to understand what it means geometrically, let's look at a common geometric shape, the circle. x2 +y2 =16 is the equation of a circle centered at the origin with radius 4. We are going to integrate xy4 along half of the circle. We first need it in parametric form. So x=4cos(t) and y=4sin(t). Keep in mind this circle will be our domain. Just like we find definite integrals over an interval, we use the curve as our analogous domain. We are going to integrate over half the circle so we need to specify t so that our curve will trace out half the circle. We can let -pi/2≤t≤pi/2 Next we compute ds so we need to find dx/dt and dy/dt. So dx/dx=-4sint and dy/dt=4cost. Now we know that ds=Square root (16sin2 t + 16cos2 t)dt=4dt. ( ds is used to show we are moving along the curve) So the integral of xy4 along a general curve c, becomes the integral of cos(t)sin4 (t) dt integrated between -pi/2 to pi/2. The result of integration along this path is 8192/5 so now we ask what does this mean geometrically? (Took a bit of time to get here, but we needed the example) Imagine putting some points on the half circle curve we used. Say point 0, 1, 2, 3, 4 etc. Now draw vectors from 0 to 1, 1 to 2, 2 to 3 etc. We have partitioned our curve into subcurves, the vectors. This is like dividing an interval into the rectangles you have often seen for Riemann integrals and Riemann sums, Instead of looking at the areas of each of those rectangles, we look at the f(pn) where pn are the points 0,1,2,3 etc mentioned above. Just like we add up the sum of the areas of the rectangles for a Riemann sum, we add up the values of the function at each point on our curve, in this case a half circle. Now if we take the limit as the vectors get smaller or as the length of the subintervals along the curve tends toward 0, this is the line integral The path is being partitioned into a polygonal path. We are letting the length or mesh or the partition go to 0. You can do this in 2 or 3 dimension space and we can integrate over a vector field. This last example helps you see an even more geometric example. Sometimes the geometry is hard to see in the examples like the one I gave above. In this one, it is pretty easy. suppose we have a function of two variables that's just a sheet above the x-y plane. Let's look at f(x,y)=5 and we want to take the line integral of this function around the closed path which is the unit circle x2+y2=1. Geometrically, the line integral is the area of a cylinder of length 1 and radius 5. You can check the this is 10Pi ( remember the surface area of a cylinder is 2Pirh, which is 2Pi5x1=10Pi) This one is easier to see geometrically because f(x,y)=5 is a sheet above the x-y plane which is easy to see. If we evaluate the that along the unit circle, then we can see how the cylinder would have to be length or height 1 and radius 5 from the sheet. To tie this to what we said earlier, think of dividing the path, the unit circle, into small subpaths or polygonal paths and evaluating f(x,y)=5 on each of these path. For each one, we get a small piece of the cylinder. If you have trouble seeing this, consider the small piece of the circle between t=0 and t=pi/1000 A very small sector. Now when we look at the value of the function on that very think sector, we have a pie shaped sector with radius 5. We add up lots of these and get the cylinder. Line integrals are also commonly used to find the work done by force long a curve. One last thing to help understand the geometry. Since we talked about vectors and dividing the curve into vectors, let's look at that geometry. Of course, the dot product of any two vectors is positive if the point in the same direction and zero if they are perpendicular. It is negative it they point in roughly opposite directions. The line integral geometrically add up dot products. If F is a vector field and DeltaR is a displacement vector. We look at the dot product of F and DeltaR long the path. If the magnitude of F is constant, the line integral will be a positive number if F is mostly pointing the same way as DeltaR and negative if F is pointing in the opposite direction. If the F is perpendicular to the path at all times, the line integral has a value of zero. We should mention one last thing ( it really is the last thing) Imagine a piece of string or wire. Let that be the curve we have been talking about analogous to the half circle etc. Now suppose we have a function f(x,y) that tells us the mass of the wire or string per unit length. The mass of the string if the line integral of f(x,y) evaluated along the string. An ordinary everyday integral is of y (a function of x) which is just a sum of all the values of the thin rectangles y times dx for all the tiny pieces of dx which make up the x axis. It can be regarded as a line integral along the line known as the x axis. But you could do an integral along any other line or curve, provided you know the value of the function at all points on the line or curve. And that is called a line integral. Instead of the fragment dx, you have a fragment ds where s is your position along the line, just as x describes where you are on the x axis.
