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Find the integral of xcos(2x) with respect to x

You can see that this question is asking you to do integration by parts. Remember that the integral of uv' is equal to uv - the integral of u'v. You want to find a u that gets easier when you differentiate i...
KJ
Answered by Krystian J. • Maths tutor
12755 Views

Integrate Sin^2(x)

(using double angle formula)Sin^2(x)=1/2-(Cos(2x)/2) So the Integral is 1/2(x-1/2Sin(2x))which simplifies to x/2 - 1/2Sin(x)Cos(x)
AR
Answered by Alex R. • Maths tutor
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Express 6cos(2x)+sin(x) in terms of sin(x). Hence solve the equation 6cos(2x) + sin(x) = 0, for 0° <= x <= 360°.

Firstly, we need to express 6cos(2x) + sin(x) in terms of sin(x) 6cos(2x) + sin(x) = 6cos(x+x) + sin(x) = 6cos 2 (x) - 6sin 2 (x) + sin(x) (applying cos(x+y) = cos(x)cos(y) - sin(x)sin(y)) = 6(1 - sin 2 (x))...
DA
Answered by Dilan A. • Maths tutor
4945 Views

A curve has equation x^2 +2xy–3y^2 +16=0. Find the coordinates of the points on the curve where dy/dx = 0.

x 2 +2xy–3y 2 +16=0Differentiate the terms:x 2 gives 2x2xy is differentiated by the product rule: vu' +v'u Make v = 2x and u = y, which gives 2x(dy/dx) + 2y3y 2 gives 6y(dy/dx)16 gives 0.Therefore we have th...
DA
Answered by Dilan A. • Maths tutor
12037 Views

The Curve C shows parametric equations x = 4tant and y = 5((3)^1/2)(sin2t) , Point P is located at (4(3)^1/2, 15/2) Find dy/dx at P.

First I would find the value of t at Point P - I would equate the x equation to 4(3)^1/2 and the y equation to 15/2. This would give me (Px,Py). After this I would then find dy/dt, and dx/dx by differentiati...
AB
Answered by Arjun B. • Maths tutor
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