Find I = ∫ tan³ x dx. The solution is: I = ½ tan² x - log cos x. * * * Here is how we can obtain this result: First, let t = tan x, s = sin x, and c = cos x; then, dI = t³ dx, ds = c dx, dc = -s dx, and dt = (1 + t²) dx; and, of course, t = s / c. By algebra, t³ = t(t² + 1) - t; thus, we have dI = t³ dx = t(t² + 1) dx - t dx = t dt - t dx. Now, d (t²) = 2t dt; thus, t dt = ½ d(t²). On the other hand, we have d log c = dc / c = -s dx / c = -t dx; thus, t dx = -d log c. Combining these results, we have dI = t dt - t dx = ½ d(t²) - d log c. This integrates readily, giving I = ½ t² - log c, which is the solution we sought. * * * We may check our result, by differentiating back: dt / dx = 1 + t²; and d(t²) / dt = 2t; thus, (d/dx)(t²) = 2t dt / dx = 2t (1 + t²). Also, we have d log c / dc = 1 / c; and dc / dx = -s; whence, (d/dx)(log c) = (dc / dx) / c = -s / c = -t. Then, dI / dx = ½ (d/dx)(t²) - (d/dx)(log c) = t (1 + t²) - t = t + t³ - t = t³, re-assuring us that we have integrated correctly.
Cotangent 32 equals tangent 0.031
Calculate the derivative of the function.Use the derivative to calculate the slope at the specified point.Calculate the y-coordinate for the point.Use the formula for a line that has a specified slope and passes through a specified point.
The integral of f'(x) = 1 is f(x) = x + c
The indefinite integral of x dt is xt
Use integration by parts: int [ln(x)] = xln(x) - int(x/x) = xln(x) - x + c
Integration for inverse tangent of square x
∫ tan(x) dx = -ln(cos(x)) + C C is the constant of integration.
∫ tahh(x) dx = ln(cosh(x)) + C C is the constant of integration.
X cubed - X cubed is zero.
An example of an education slogan for India is "India is tangent 45,varieties of people are tangent 90 but enemity is tangent 0" A slogan for general purposes is "Education for national integration" or "National integration through education."
To integrate tan(x), you must break up tangent into sine over cosine, with that being done, all you have is a u-substitution with the cosine. This should give: int(tan(x)dx)=int(sin(x)/cos(x)dx)=int(-(1/u)*du)=-ln|u|+C=-ln|cos(x)|+C u=cos(x) du=-sin(x)dx
X 2 cubed
x to the power of 6.
x times x times x equals x cubed
x6
it is x to the sixth power
Well, since a tangent line touches a line in one spot, the Y axis could be considered tangent to the X axis.