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13 Nov 2019
I keep getting this wrong
Tutorial Exercise Evaluate the indefinite integral x36 sin(x37) dx Step 1 We must decide what to choose for u. If u = rx), then du = f (x) dx, and so it is helpful to look for some expression in which the derivative is also present, though perhaps missing a constant factor. For example, x37 is part of this integral, and the derivative of x37 is 37.36 also present except for a constant. x36 sin(x37) dx for which is 37r36 Step 2 If we choose u=x37, then du=37x36 dx. If u = x37 is substituted into J x36 sin(x37) dx- then we have J x36 sin ud,-J sin u(x36 dx). We must also convert x36 dx into an expression involving u. We know that du = 37x36 dx, and so x36 dx = Submit Skip (ou cannot.com x du Skip (vou cannot come back
I keep getting this wrong
Tutorial Exercise Evaluate the indefinite integral x36 sin(x37) dx Step 1 We must decide what to choose for u. If u = rx), then du = f (x) dx, and so it is helpful to look for some expression in which the derivative is also present, though perhaps missing a constant factor. For example, x37 is part of this integral, and the derivative of x37 is 37.36 also present except for a constant. x36 sin(x37) dx for which is 37r36 Step 2 If we choose u=x37, then du=37x36 dx. If u = x37 is substituted into J x36 sin(x37) dx- then we have J x36 sin ud,-J sin u(x36 dx). We must also convert x36 dx into an expression involving u. We know that du = 37x36 dx, and so x36 dx = Submit Skip (ou cannot.com x du Skip (vou cannot come back
Bunny GreenfelderLv2
2 Nov 2019