Use integration to find I = ∫ xsin3x dx
Use integration by parts, let U = x, the derivative of U = 1, let the derivative of V = sin3x and intergrate the derivative of V to arrive at V = (-1/3)(cos3x). Substitute the value into the formula uv − ∫ vdu dx dx, arrive at I = (x)(-1/3)(cos3x) - ∫(1)(-1/3)(cos3x)dx which can be written us I = (-x/3)(cos3x) +∫(1/3)(cos3x)dx. ∫(1)(1/3)(cos3x)dx = (1/9)(sin3x). Now put that into the original equation giving the final answer I = (-x/3)(cos3x)+ (1/9)(sin3x) + c,
ZL
Related Maths A Level answers
All answers ▸Consider the closed curve between 0 <= theta < 2pi given by r(theta) = 6 + alpha sin theta, where alpha is some real constant strictly between 0 and 6. The area in this closed curve is 97pi/2. Calculate the value of alpha.
Solve 2sec^2(x) = 3 + tan(x) for 0 < x <pi/2
How do we differentiate y=a^x when 'a' is an non zero real number
At time t = 0, a particle is projected vertically upwards with speed u m s–1 from a point 10 m above the ground. At time T seconds, the particle hits the ground with speed 17.5 m s–1. Find the value of u and T and evaluate the model. (AS mechanics)