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What is the indefinite integral of 3sinx-5cosx?
8
What is the integral for sin3x?
S sin3X dx =-1/3 cos3X
Integral of x cosx dx?
The integral of x cos(x) dx is cos(x) + x sin(x) + C
How do you Differentiate and Integrate y equals 3x?
dy/dx = 3 integral = (3x^2)/2
What is the integral of tan x-3 dx?
In this specific example one would need to use the u substitution method. * Set u to be x - 3 * Derive x - 3 * u = x - 3 * du = dx Now that we have integrated u we can remove the x - 3 and substitute in u and remove the dx and substitute in du. This is what we have after substituting: * (the integrand of) tan(u)du Now integrate tan(u)du * the Integral of tan(u)du is: * sec2(u) Now resubstitute what we set as u. In this case we set x - 3 to u. This will give us our final answer and integral of tan(x-3)dx. * sec2(x - 3)
What is the indefinite integral?
An indefinite integral is a version of an integral that, unlike a definite integral, returns an expression instead of a number. The general form of a definite integral is: ∫ba f(x) dx. The general form of an indefinite integral is: ∫ f(x) dx. An example of a definite integral is: ∫20 x2 dx. An example of an indefinite integral is: ∫ x2 dx In the definite case, the answer is 23/3 - 03/3 = 8/3. In the indefinite case, the answer is x3/3 + C, where C is an arbitrary constant.
What is the integral of -3dx?
∫(-3)dx = -3x + C
What is the indefinite integral of 3sinx-5cosx?
8
What is the integral for sin3x?
S sin3X dx =-1/3 cos3X
Integral of 1 divided by sinx cosx?
Integral of [1/(sin x cos x) dx] (substitute sin2 x + cos2 x for 1)= Integral of [(sin2 x + cos2 x)/(sin x cos x) dx]= Integral of [sin2 x/(sin x cos x) dx] + Integral of [cos2 x/(sin x cos x) dx]= Integral of (sin x/cos x dx) + Integral of (cos x/sin x dx)= Integral of tan x dx + Integral of cot x dx= ln |sec x| + ln |sin x| + C
Integration of tan pow4x?
integral of (tanx)^4 (tanx)^4 = (tanx)^2 (tanx)^2 =(sec^2 x - 1)(tan^2 x) =(sec^2 x)(tan^2 x) - tan^2 x = integral of sec^2 x tan^2 x dx - integral of tan^2 x dx First, integral of sec^2 x tan^2 x dx Let u = tanx because that would make du = sec^2 x dx so then we have integral of u^2 du which is (1/3)u^3 substituting back in tanx we get (1/3)tan^3 x Next, integral of tan^2 x tan^2 x = sec^2 x -1 integral of sec^2 x - 1 = integral of sec^2 x dx - integral 1 dx = tanx - x so putting it all together we have integral of tan^4 x dx = (1/3)tan^3 x - tanx + x + C
Integral of x cosx dx?
The integral of x cos(x) dx is cos(x) + x sin(x) + C
Integral formula of a raised to x?
integral (a^x) dx = (a^x) / ln(a)
What is the integral from 0 to pi over 6 sine 2x dx?
Integral from 0 to pi 6sin2xdx: integral of 6sin2xdx (-3)cos2x+c. (-3)cos(2 x pi) - (-3)cos(2 x 0) -3 - -3 0
What is the integral of -3x?
In order to compute that integral, we need to use the power rule: ∫ xⁿ dx = xn + 1/(n + 1) + c where n is any constant except 0 and -1. Apply that rule to get: ∫ 3x dx = 3 ∫ x dx [Factor out the constant] = 3 ∫ x1 dx [Make note of the exponent] = 3x1 + 1/(1 + 1) + c = 3x2/2 + c So that is the integral of 3x.
What is the integral of x divided by sq-rt of x plus 2?
Integral of x dx / sqrt(x+2) Make the substitution sqrt(x+2)=u (x+2)^(1/2) = u (1/2)(x+2)^(-1/2) dx = du 1/2(x+2)^(1/2) dx = du 1/2sqrt(x+2) dx = du 1/sqrt(x+2) dx = 2 du Integral of x dx / sqrt(x+2) = Integral 2 x du sqrt(x+2) = u (x+2)=u^2 x=u^2-2 Integral 2 x du = Integral 2(u^2-2) du = Integral 2u^2 du - 4 du = 2 u^3/3 - 4u + C = (2/3) (x+2)^(3/2) - 4 sqrt(x+2) + C
How do you Differentiate and Integrate y equals 3x?
dy/dx = 3 integral = (3x^2)/2
