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Indefinite Integrals

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rebel

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rebelmaths

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$\\int(x^4-\\var{a}x^3+\\var{b}x-\\var{c})\\mathrm{dx}$

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$\\int(u+\\var{d})(2u+\\var{f})\\mathrm{du}$

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$\\int\\frac{\\sin(2x)}{\\sin(x)}\\mathrm{dx}$.

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Take care to include the brackets in the trigonometric expressions, i.e. write sin(x) rather than sinx.

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Solve the following indefinite integrals, using $C$ to represent an unknown constant.

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(a)

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$\\int(x^4-\\var{a}x^3+\\var{b}x-\\var{c})\\mathrm{dx}$

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Using $\\int \\;x^n\\;dx=\\frac{x^{n+1}}{n+1}+C$ we obtain

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$\\int(x^4-\\var{a}x^3+\\var{b}x-\\var{c})\\mathrm{dx} = \\simplify{x^5/5-{a/4}x^4+{b/2}x^2-{c}x+C}$

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(b)

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To find $\\int(u+\\var{d})(2u+\\var{f})\\mathrm{du}$ we can first multiply out the brackets to obtain:

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$\\int \\simplify{2u^2+{2d+f}u+{d*f}}\\;{du} = \\simplify{2/3u^3+{(2d+f)/2}u^2+{d*f}u+C} $

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(c)

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To find $\\int\\frac{\\sin(2x)}{\\sin(x)}\\mathrm{dx}$ we use the identity $\\sin(2x)=2\\sin(x)\\cos(x)$

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Hence our integral becomes $\\int\\frac{2\\sin(x)\\cos(x)}{\\sin(x)}\\mathrm{dx}=\\int2\\cos(x)\\mathrm{dx}=2\\sin(x)+C$

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