Using the various versions of $\\cos{2x}$ identity to integrate $\\sin^2{x}$ and $\\cos^2{x}$.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Integrate $f(x)=\\var{Func}$.
", "advice": "We can't integrate $\\var{Coeff}\\sin^2(x)$ directly so first we have to use the double angle formula $\\cos(2x)=1-2\\sin^2(x)$. We re-arrange using the double angle formula to give us,
\n\\[\\var{Coeff}\\sin^2(x)=\\frac{\\var{Coeff}}2-\\frac{\\var{Coeff}}2\\cos(2x).\\]
\nFrom the Table of Integrals we see that a function of the form \\[ f(x)= \\cos(nx) \\] has the integral \\[ \\int \\cos(nx) dx = \\frac{1}{n}\\sin(nx)+c\\]
\n\nSo, for the function
\n\\[f(x)=\\simplify[unitFactor,fractionNumbers]{{-Coeff/2}cos(2x)},\\]
\nthe integral is
\n\\[ \\begin{split} \\int\\simplify[unitFactor,fractionNumbers]{{-Coeff/2}cos(2x)} dx \\,= \\simplify[unitFactor,fractionNumbers]{{-Coeff/2}int(cos(2x),x)} &\\,=\\simplify[unitFactor,fractionNumbers]{{-Coeff/2}(1/2 sin({2}x))} +c, \\\\ &\\,=\\simplify[unitFactor,fractionNumbers]{{-Coeff/4} sin(2x)+c}. \\end{split} \\]
\nThe integral of $\\frac{\\var{Coeff}}2$ is
\n\\[\\int\\frac{\\var{Coeff}}2dx=\\frac{\\var{Coeff}}2x+c,\\]
\nso combining these our final answer is
\n\\[\\int\\frac{\\var{Coeff}}2-\\frac{\\var{Coeff}}2\\cos(2x)dx=\\simplify[unitFactor,fractionNumbers]{{Coeff/2}x-{Coeff/4} sin(2x)+c}\\]
\nWe can't integrate $\\var{Coeff}\\cos^2(x)$ directly so first we have to use the double angle formula $\\cos(2x)=2\\cos^2(x)-1$. We re-arrange using the double angle formula to give us,
\n\\[\\var{Coeff}\\cos^2(x)=\\frac{\\var{Coeff}}2+\\frac{\\var{Coeff}}2\\cos(2x).\\]
\nFrom the Table of Integrals we see that a function of the form \\[ f(x)= \\cos(nx) \\] has the integral \\[ \\int \\cos(nx) dx = \\frac{1}{n}\\sin(nx)+c\\]
\nSo, for the function
\n\\[f(x)=\\simplify[unitFactor,fractionNumbers]{{Coeff/2}cos(2x)},\\]
\nthe integral is
\n\\[ \\begin{split} \\int\\simplify[unitFactor,fractionNumbers]{{Coeff/2}cos(2x)} dx \\,= \\simplify[unitFactor,fractionNumbers]{{Coeff/2}int(cos(2x),x)} &\\,=\\simplify[unitFactor,fractionNumbers]{{Coeff/2}(1/2 sin({2}x))} +c, \\\\ &\\,=\\simplify[unitFactor,fractionNumbers]{{Coeff/4} sin(2x)+c}. \\end{split} \\]
\nThe integral of $\\frac{\\var{Coeff}}2$ is
\n\\[\\int\\frac{\\var{Coeff}}2dx=\\frac{\\var{Coeff}}2x+c,\\]
\nso combining these our final answer is
\n\\[\\int\\frac{\\var{Coeff}}2+\\frac{\\var{Coeff}}2\\cos(2x)dx=\\simplify[unitFactor,fractionNumbers]{{Coeff/2}x+{Coeff/4} sin(2x)+c}\\]
Use this link to find some resources which will help you revise this topic.
It looks like you forgot to include the integration constant. You should always remember the \"+C\" when doing an indefinite integral.
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