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Firstly let's take a right angle triangle. Label its sides a,b, c for the hypotenuse. Then arrange 4 of these triangles into a large square, with a smaller square contained within it. The area of the smal...
Integrating this expression is a simple trick. We use integration by parts. For this we need a function we can integrate and a function we can differentiate. We know how to differentiate ln(x) which is 1/...
We can start with the identity sin2(x)+cos2(x)=1 If we divide through the equation by cos2(x), we get: sin2(x)/cos2(x) + cos2(x)/cos...
To get the required tangent equation we need its gradient and the coordinates of a point it passes through. We can then substitute it into the formula y - y1 = m(x - x1). 1. ...
You use integration by parts if there's more than one function of x. For example; to integrate xsin(x) you'd use integration by parts as this is two functions of x. The formula to remember is I(udv) =...
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