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Are we comfortable with an integral being the area under a curve on a graph? Well we'll draw a curve and think about how we could estimate this area. Would you agree that we could chop up the area under t...
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 differentiat...
(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)
Firstly, we need to express 6cos(2x) + sin(x) in terms of sin(x) 6cos(2x) + sin(x) = 6cos(x+x) + sin(x) = 6cos2(x) - 6sin2(x) + sin(x) (applying cos(x+y) = cos(x)cos(y) - sin(x)sin(y...
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