integral from 0 to 1 of 9cos((pit)/2)

{ "query": { "display": "$$\\int_{0}^{1}9\\cos\\left(\\frac{πt}{2}\\right)dt$$", "symbolab_question": "BIG_OPERATOR#\\int _{0}^{1}9\\cos(\\frac{πt}{2})dt" }, "solution": { "level": "PERFORMED", "subject": "Calculus", "topic": "Integrals", "subTopic": "Definite Integrals", "default": "\\frac{18}{π}", "decimal": "5.72957…", "meta": { "showVerify": true } }, "steps": { "type": "interim", "title": "$$\\int_{0}^{1}9\\cos\\left(\\frac{πt}{2}\\right)dt=\\frac{18}{π}$$", "input": "\\int_{0}^{1}9\\cos\\left(\\frac{πt}{2}\\right)dt", "steps": [ { "type": "step", "primary": "Take the constant out: $$\\int{a\\cdot{f\\left(x\\right)}dx}=a\\cdot\\int{f\\left(x\\right)dx}$$", "result": "=9\\cdot\\:\\int_{0}^{1}\\cos\\left(\\frac{πt}{2}\\right)dt" }, { "type": "interim", "title": "Apply u-substitution", "input": "\\int_{0}^{1}\\cos\\left(\\frac{πt}{2}\\right)dt", "steps": [ { "type": "definition", "title": "Integral Substitution definition", "text": "$$\\int\\:f\\left(g\\left(x\\right)\\right)\\cdot\\:g'\\left(x\\right)dx=\\int\\:f\\left(u\\right)du,\\:\\quad\\:u=g\\left(x\\right)$$", "secondary": [ "Substitute: $$u=\\frac{πt}{2}$$" ] }, { "type": "interim", "title": "$$\\frac{du}{dt}=\\frac{π}{2}$$", "input": "\\frac{d}{dt}\\left(\\frac{πt}{2}\\right)", "steps": [ { "type": "step", "primary": "Take the constant out: $$\\left(a{\\cdot}f\\right)'=a{\\cdot}f'$$", "result": "=\\frac{π}{2}\\frac{dt}{dt}" }, { "type": "step", "primary": "Apply the common derivative: $$\\frac{dt}{dt}=1$$", "result": "=\\frac{π}{2}\\cdot\\:1" }, { "type": "step", "primary": "Simplify", "result": "=\\frac{π}{2}", "meta": { "solvingClass": "Solver" } } ], "meta": { "solvingClass": "Derivatives", "interimType": "Derivatives", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s79Kg+idP5vLVrjUll6eMdYiT0VcbSpgSwaS31VE8/n/hZUSejOWlsBZ/+JGlie2TqQ5BSVQv3cb+dhTKK218l6GYLiktW9nIHNLNg9B2jDC//P/+v51eTuH2/F4MIu5mq3tSj8g1LlTnRuF1rA2RzPdW/IMN/06+UDp77WW8WfQs=" } }, { "type": "step", "primary": "$$\\quad\\Rightarrow\\:du=\\frac{π}{2}dt$$" }, { "type": "step", "primary": "$$\\quad\\Rightarrow\\:dt=\\frac{2}{π}du$$" }, { "type": "step", "result": "=\\int\\:\\cos\\left(u\\right)\\frac{2}{π}du" }, { "type": "step", "primary": "Adjust integral boundaries:" }, { "type": "interim", "title": "$$t=0\\quad\\Rightarrow\\:u=0$$", "input": "u=\\frac{πt}{2}", "steps": [ { "type": "step", "primary": "Plug in $$t=0$$", "result": "=\\frac{π0}{2}" }, { "type": "step", "primary": "Apply rule $$0\\cdot\\:a=0$$", "result": "=\\frac{0}{2}" }, { "type": "step", "primary": "Apply rule $$\\frac{0}{a}=0,\\:a\\ne\\:0$$", "result": "=0" } ], "meta": { "solvingClass": "Solver", "interimType": "Solver", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s73EjSnH3l2ufT1DqTo5o77lnyYRz18HvB+rp63mPitc8E5aqGN/sLZfeoFZRwtGLqP8vQyhiD4JSfqjIvcQ7tip9r8Rk0z4iQ05WW3sAcDOsriOVvxdxzDZmtIBCXLSv+" } }, { "type": "interim", "title": "$$t=1\\quad\\Rightarrow\\:u=\\frac{π}{2}$$", "input": "u=\\frac{πt}{2}", "steps": [ { "type": "step", "primary": "Plug in $$t=1$$", "result": "=\\frac{π1}{2}" }, { "type": "step", "primary": "Multiply: $$π1=π$$", "result": "=\\frac{π}{2}" } ], "meta": { "solvingClass": "Solver", "interimType": "Solver", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s73EjSnH3l2ufT1DqTo5o77lnyYRz18HvB+rp63mPitc/9ovYKijQYhJDCbxu/nAOJvbezqhqvja4IrwPu6KhhTnbn58iI+3Qqf4c2/hnjL5XfHgF3Nh2mtUhcGjElkadVGSx3zIWTEe9Xds4zs4JG7ImpXFf3SOUx+H18qfp3MLg=" } }, { "type": "step", "result": "=\\int_{0}^{\\frac{π}{2}}\\cos\\left(u\\right)\\frac{2}{π}du" } ], "meta": { "interimType": "Integral U Substitution 1Eq", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7l7+Hp9GxvB/5I+4/5L7s7++2f7fkPjY0fx3zeZzGzu8DUdNpLSYnseysobGCxLZvk3WldPTzMRCmfRYnoIUxcJOjCbeNi4QsIc/hhk+abxaltxFanESN+CNruhhOhs3JkiCRTCBASun8z7Ea1vSIjqAdjzj4NsO1VbHrGBYsiKB5kCzK3JoeaJR362EmWApGUU0ZlAG+PwFBTdWQXx/Mp0rbw/BigzSSTfKortF0XFWVANwMfaMnAbc2ZySQFC5FacUWHcZ12i9G+tzuywix7A==" } }, { "type": "step", "result": "=9\\cdot\\:\\int_{0}^{\\frac{π}{2}}\\cos\\left(u\\right)\\frac{2}{π}du" }, { "type": "step", "primary": "Take the constant out: $$\\int{a\\cdot{f\\left(x\\right)}dx}=a\\cdot\\int{f\\left(x\\right)dx}$$", "result": "=9\\cdot\\:\\frac{2}{π}\\cdot\\:\\int_{0}^{\\frac{π}{2}}\\cos\\left(u\\right)du" }, { "type": "step", "primary": "Use the common integral: $$\\int\\:\\cos\\left(u\\right)du=\\sin\\left(u\\right)$$", "result": "=9\\cdot\\:\\frac{2}{π}[\\sin\\left(u\\right)]_{0}^{\\frac{π}{2}}" }, { "type": "interim", "title": "Simplify $$9\\cdot\\:\\frac{2}{π}[\\sin\\left(u\\right)]_{0}^{\\frac{π}{2}}:{\\quad}\\frac{18}{π}[\\sin\\left(u\\right)]_{0}^{\\frac{π}{2}}$$", "input": "9\\cdot\\:\\frac{2}{π}[\\sin\\left(u\\right)]_{0}^{\\frac{π}{2}}", "result": "=\\frac{18}{π}[\\sin\\left(u\\right)]_{0}^{\\frac{π}{2}}", "steps": [ { "type": "step", "primary": "Multiply fractions: $$a\\cdot\\frac{b}{c}=\\frac{a\\:\\cdot\\:b}{c}$$", "result": "=\\frac{2\\cdot\\:9}{π}[\\sin\\left(u\\right)]_{0}^{\\frac{π}{2}}" }, { "type": "step", "primary": "Multiply the numbers: $$2\\cdot\\:9=18$$", "result": "=\\frac{18}{π}[\\sin\\left(u\\right)]_{0}^{\\frac{π}{2}}" } ], "meta": { "solvingClass": "Solver", "interimType": "Generic Simplify Specific 1Eq", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7tCaovvy4r4yHTcJnXLMZMObkXe2au5RDPGdIM+hdNJBJQwXhEh4xXpe/PwugporPqSLiRo0DO9IhVuX5Gm3ZK3CQoYlYQ8U+Tfyx0kyzI8hEYup6E89obXB1653chS0A/5khMjBBaStwKte1COzYi2rIShxvg95bTrojH9Wv/TSjeh7+jKEzLb7VNCEMF3Z/bMzoTd+5nEXVeQoBhpFcICE4CAPxzUGxk4xstK5chykb6ghkjx/zVwkoG++QZFiu8V+pjA3MzMgzCNUDlrU9p1iVI3uvN1by+AN9NfjoKFU=" } }, { "type": "interim", "title": "Compute the boundaries:$${\\quad}1$$", "input": "[\\sin\\left(u\\right)]_{0}^{\\frac{π}{2}}", "steps": [ { "type": "step", "primary": "$$\\int_{a}^{b}{f\\left(x\\right)dx}=F\\left(b\\right)-F\\left(a\\right)=\\lim_{x\\to\\:b-}\\left(F\\left(x\\right)\\right)-\\lim_{x\\to\\:a+}\\left(F\\left(x\\right)\\right)$$" }, { "type": "interim", "title": "$$\\lim_{u\\to\\:0+}\\left(\\sin\\left(u\\right)\\right)=0$$", "input": "\\lim_{u\\to\\:0+}\\left(\\sin\\left(u\\right)\\right)", "steps": [ { "type": "step", "primary": "Plug in the value $$u=0$$", "result": "=\\sin\\left(0\\right)", "meta": { "title": { "extension": "Limit properties - if the limit of f(x), and g(x) exists, then:<br/>$$\\bullet\\quad\\lim_{x\\to\\:a}\\left(x\\right)=a$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}[c\\cdot{f\\left(x\\right)}]=c\\cdot\\lim_{x\\to{a}}{f\\left(x\\right)}$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}[\\left(f\\left(x\\right)\\right)^c]=\\left(\\lim_{x\\to{a}}{f\\left(x\\right)}\\right)^c$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}[f\\left(x\\right)\\pm{g\\left(x\\right)}]=\\lim_{x\\to{a}}{f\\left(x\\right)}\\pm\\lim_{x\\to{a}}{g\\left(x\\right)}$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}[f\\left(x\\right)\\cdot{g\\left(x\\right)}]=\\lim_{x\\to{a}}{f\\left(x\\right)}\\cdot\\lim_{x\\to{a}}{g\\left(x\\right)}$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}\\left(\\frac{f\\left(x\\right)}{g\\left(x\\right)}\\right)=\\frac{\\lim_{x\\to{a}}{f\\left(x\\right)}}{\\lim_{x\\to{a}}{g\\left(x\\right)}},\\:$$where $$\\lim_{x\\to{a}}g\\left(x\\right)\\neq0$$" } } }, { "type": "step", "primary": "Use the following trivial identity:$${\\quad}\\sin\\left(0\\right)=0$$", "secondary": [ "$$\\sin\\left(x\\right)$$ periodicity table with $$2πn$$ cycle:<br/>$$\\begin{array}{|c|c|c|c|}\\hline x&\\sin(x)&x&\\sin(x)\\\\\\hline 0&0&π&0\\\\\\hline \\frac{π}{6}&\\frac{1}{2}&\\frac{7π}{6}&-\\frac{1}{2}\\\\\\hline \\frac{π}{4}&\\frac{\\sqrt{2}}{2}&\\frac{5π}{4}&-\\frac{\\sqrt{2}}{2}\\\\\\hline \\frac{π}{3}&\\frac{\\sqrt{3}}{2}&\\frac{4π}{3}&-\\frac{\\sqrt{3}}{2}\\\\\\hline \\frac{π}{2}&1&\\frac{3π}{2}&-1\\\\\\hline \\frac{2π}{3}&\\frac{\\sqrt{3}}{2}&\\frac{5π}{3}&-\\frac{\\sqrt{3}}{2}\\\\\\hline \\frac{3π}{4}&\\frac{\\sqrt{2}}{2}&\\frac{7π}{4}&-\\frac{\\sqrt{2}}{2}\\\\\\hline \\frac{5π}{6}&\\frac{1}{2}&\\frac{11π}{6}&-\\frac{1}{2}\\\\\\hline \\end{array}$$" ], "result": "=0", "meta": { "solvingClass": "Solver" } } ], "meta": { "solvingClass": "Limits", "interimType": "Limits" } }, { "type": "interim", "title": "$$\\lim_{u\\to\\:\\frac{π}{2}-}\\left(\\sin\\left(u\\right)\\right)=1$$", "input": "\\lim_{u\\to\\:\\frac{π}{2}-}\\left(\\sin\\left(u\\right)\\right)", "steps": [ { "type": "step", "primary": "Plug in the value $$u=\\frac{π}{2}$$", "result": "=\\sin\\left(\\frac{π}{2}\\right)", "meta": { "title": { "extension": "Limit properties - if the limit of f(x), and g(x) exists, then:<br/>$$\\bullet\\quad\\lim_{x\\to\\:a}\\left(x\\right)=a$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}[c\\cdot{f\\left(x\\right)}]=c\\cdot\\lim_{x\\to{a}}{f\\left(x\\right)}$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}[\\left(f\\left(x\\right)\\right)^c]=\\left(\\lim_{x\\to{a}}{f\\left(x\\right)}\\right)^c$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}[f\\left(x\\right)\\pm{g\\left(x\\right)}]=\\lim_{x\\to{a}}{f\\left(x\\right)}\\pm\\lim_{x\\to{a}}{g\\left(x\\right)}$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}[f\\left(x\\right)\\cdot{g\\left(x\\right)}]=\\lim_{x\\to{a}}{f\\left(x\\right)}\\cdot\\lim_{x\\to{a}}{g\\left(x\\right)}$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}\\left(\\frac{f\\left(x\\right)}{g\\left(x\\right)}\\right)=\\frac{\\lim_{x\\to{a}}{f\\left(x\\right)}}{\\lim_{x\\to{a}}{g\\left(x\\right)}},\\:$$where $$\\lim_{x\\to{a}}g\\left(x\\right)\\neq0$$" } } }, { "type": "step", "primary": "Use the following trivial identity:$${\\quad}\\sin\\left(\\frac{π}{2}\\right)=1$$", "secondary": [ "$$\\sin\\left(x\\right)$$ periodicity table with $$2πn$$ cycle:<br/>$$\\begin{array}{|c|c|c|c|}\\hline x&\\sin(x)&x&\\sin(x)\\\\\\hline 0&0&π&0\\\\\\hline \\frac{π}{6}&\\frac{1}{2}&\\frac{7π}{6}&-\\frac{1}{2}\\\\\\hline \\frac{π}{4}&\\frac{\\sqrt{2}}{2}&\\frac{5π}{4}&-\\frac{\\sqrt{2}}{2}\\\\\\hline \\frac{π}{3}&\\frac{\\sqrt{3}}{2}&\\frac{4π}{3}&-\\frac{\\sqrt{3}}{2}\\\\\\hline \\frac{π}{2}&1&\\frac{3π}{2}&-1\\\\\\hline \\frac{2π}{3}&\\frac{\\sqrt{3}}{2}&\\frac{5π}{3}&-\\frac{\\sqrt{3}}{2}\\\\\\hline \\frac{3π}{4}&\\frac{\\sqrt{2}}{2}&\\frac{7π}{4}&-\\frac{\\sqrt{2}}{2}\\\\\\hline \\frac{5π}{6}&\\frac{1}{2}&\\frac{11π}{6}&-\\frac{1}{2}\\\\\\hline \\end{array}$$" ], "result": "=1", "meta": { "solvingClass": "Solver" } } ], "meta": { "solvingClass": "Limits", "interimType": "Limits" } }, { "type": "step", "result": "=1-0" }, { "type": "step", "primary": "Simplify", "result": "=1", "meta": { "solvingClass": "Solver" } } ], "meta": { "interimType": "Integral Definite Limit Boundaries 0Eq", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7l7+Hp9GxvB/5I+4/5L7s74ULt1msigtcOhDTdf61DSE5noRtMY594oojBol9JwC9XfA9cUywaDbEKRKQOwovlKLmtAwy2Lt01l2BnPhHmpi3gYcJWoW3DPkUSnPpO877dEoWy0Hj9x/Ad2N4oq/YXvC30sSftAIFS6Qkpy19Ikqv3jTgg7FSw5xBGGG9m4x1mC7WtKAr21NimcVlxHR1cA==" } }, { "type": "step", "result": "=\\frac{18}{π}\\cdot\\:1" }, { "type": "step", "primary": "Simplify", "result": "=\\frac{18}{π}" } ], "meta": { "solvingClass": "Integrals" } }, "plot_output": { "meta": { "plotInfo": { "variable": "t", "plotRequest": "yes" }, "showViewLarger": true } }, "meta": { "showVerify": true } }