integral from 0 to pi/4 of cot(x)

{ "query": { "display": "$$\\int_{0}^{\\frac{π}{4}}\\cot\\left(x\\right)dx$$", "symbolab_question": "BIG_OPERATOR#\\int _{0}^{\\frac{π}{4}}\\cot(x)dx" }, "solution": { "level": "PERFORMED", "subject": "Calculus", "topic": "Integrals", "subTopic": "Definite Integrals", "default": "\\mathrm{diverges}", "meta": { "showVerify": true } }, "steps": { "type": "interim", "title": "$$\\int_{0}^{\\frac{π}{4}}\\cot\\left(x\\right)dx=$$diverges", "input": "\\int_{0}^{\\frac{π}{4}}\\cot\\left(x\\right)dx", "steps": [ { "type": "interim", "title": "Rewrite using trig identities", "input": "\\int_{0}^{\\frac{π}{4}}\\cot\\left(x\\right)dx", "result": "=\\int_{0}^{\\frac{π}{4}}\\frac{\\cos\\left(x\\right)}{\\sin\\left(x\\right)}dx", "steps": [ { "type": "step", "primary": "Use the following identity: $$\\cot\\left(x\\right)=\\frac{\\cos\\left(x\\right)}{\\sin\\left(x\\right)}$$", "result": "=\\int_{0}^{\\frac{π}{4}}\\frac{\\cos\\left(x\\right)}{\\sin\\left(x\\right)}dx" } ], "meta": { "interimType": "Trig Rewrite Using Trig identities 0Eq", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7l7+Hp9GxvB/5I+4/5L7s7/FfqYwNzMzIMwjVA5a1PadPrZr1YEt17AAaJ6hxcb2W9X1rrZQlOM1M4JBXIVFZSjnloLBODJONKaxA2ZCkPF3KuuPm/GILl3+KgrSlvW+yRYcUszXI4D819dCO5582z9LIdjDoqrKCC/TfWDltBS5KzexUGP1Jw3gZceaq16SWWCbBHh/M5vnCPTIJW0J20psckulJqWLpUQLN4Of4cBYYcvEp+ArJ7IgQQisySYDH4va4o6ZZQt6EK5Zci1d56K6g7caZnrJReWEr4vU9em8=" } }, { "type": "interim", "title": "Apply u-substitution", "input": "\\int_{0}^{\\frac{π}{4}}\\frac{\\cos\\left(x\\right)}{\\sin\\left(x\\right)}dx", "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: $$u=\\sin\\left(x\\right)$$" ] }, { "type": "interim", "title": "$$\\frac{du}{dx}=\\cos\\left(x\\right)$$", "input": "\\frac{d}{dx}\\left(\\sin\\left(x\\right)\\right)", "steps": [ { "type": "step", "primary": "Apply the common derivative: $$\\frac{d}{dx}\\left(\\sin\\left(x\\right)\\right)=\\cos\\left(x\\right)$$", "result": "=\\cos\\left(x\\right)" } ], "meta": { "solvingClass": "Derivatives", "interimType": "Derivatives", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s79Kg+idP5vLVrjUll6eMdYgOt2FhQQwx0GxLGzv2mPOv8zeERICEnv1Ds5A1/BdIwQslTDKxOR/6J+ZOGvUcaugB66mSUqneplfTkjggryzD6iDpcTVxDjQ5tzND5SL/a/YPuLKiL1T3raVYWYuVCmA==" } }, { "type": "step", "primary": "$$\\quad\\Rightarrow\\:du=\\cos\\left(x\\right)dx$$" }, { "type": "step", "primary": "$$\\quad\\Rightarrow\\:dx=\\frac{1}{\\cos\\left(x\\right)}du$$" }, { "type": "step", "result": "=\\int\\:\\frac{\\cos\\left(x\\right)}{u}\\cdot\\:\\frac{1}{\\cos\\left(x\\right)}du" }, { "type": "interim", "title": "Simplify $$\\frac{\\cos\\left(x\\right)}{u}\\cdot\\:\\frac{1}{\\cos\\left(x\\right)}:{\\quad}\\frac{1}{u}$$", "input": "\\frac{\\cos\\left(x\\right)}{u}\\cdot\\:\\frac{1}{\\cos\\left(x\\right)}", "steps": [ { "type": "step", "primary": "Multiply fractions: $$\\frac{a}{b}\\cdot\\frac{c}{d}=\\frac{a\\:\\cdot\\:c}{b\\:\\cdot\\:d}$$", "result": "=\\frac{\\cos\\left(x\\right)\\cdot\\:1}{u\\cos\\left(x\\right)}" }, { "type": "step", "primary": "Cancel the common factor: $$\\cos\\left(x\\right)$$", "result": "=\\frac{1}{u}" } ], "meta": { "solvingClass": "Solver", "interimType": "Algebraic Manipulation Simplify Title 1Eq" } }, { "type": "step", "result": "=\\int\\:\\frac{1}{u}du" }, { "type": "step", "primary": "Adjust integral boundaries:" }, { "type": "interim", "title": "$$x=0\\quad\\Rightarrow\\:u=0$$", "input": "u=\\sin\\left(x\\right)", "steps": [ { "type": "step", "primary": "Plug in $$x=0$$", "result": "=\\sin\\left(0\\right)" }, { "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", "interimType": "Solver", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7iz+Lh9X3qTzAkE1p7nYhzSAn9lkDfZkicUGkO3EF+Ip/suoH0JzK9gJYOwax5fT2MkoyjE7DBeWY9MXUJBbE1lUUew5fVKmjQ3z3Sfxbm0Ax5n8cqw2/G+CGXoIe1phy" } }, { "type": "interim", "title": "$$x=\\frac{π}{4}\\quad\\Rightarrow\\:u=\\frac{\\sqrt{2}}{2}$$", "input": "u=\\sin\\left(x\\right)", "steps": [ { "type": "step", "primary": "Plug in $$x=\\frac{π}{4}$$", "result": "=\\sin\\left(\\frac{π}{4}\\right)" }, { "type": "step", "primary": "Use the following trivial identity:$${\\quad}\\sin\\left(\\frac{π}{4}\\right)=\\frac{\\sqrt{2}}{2}$$", "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": "=\\frac{\\sqrt{2}}{2}" } ], "meta": { "solvingClass": "Solver", "interimType": "Solver", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7iz+Lh9X3qTzAkE1p7nYhzSAn9lkDfZkicUGkO3EF+IpdgYKaIEX32V6yZdn2Ih8ns2lAxHJPmDooQNuNmhZntC3YgxvxZy+FYwpFRnQjqKrdWxV5EMZ1eajXZFHxuGATXAqbIz4se2xKhMuUBfup5rHvFY0gMKcgzk7LDCp2tDTkfreZAsHeOkLcfJ1ekMZT" } }, { "type": "step", "result": "=\\int_{0}^{\\frac{\\sqrt{2}}{2}}\\frac{1}{u}du" } ], "meta": { "interimType": "Integral U Substitution 1Eq", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7l7+Hp9GxvB/5I+4/5L7s7/FfqYwNzMzIMwjVA5a1PacHMqn7G2tLilAbTq7aKmXXpi/4sxAX6eK685hGvPSvNhyOUtErWNRXbTqqjj69E3vJGcpJEkEeycRerXBL/PDDuHwpRf2E3bXAkTXof819nCOlPIX71ZKkUigvXkOYNVGeFN1UF1AUTUJmofrq1Ro9gNLsLN1hP1mx9E83xuA+wf8DOFkEB41JAUo0zfus0lAEuDOVaQvKofqHoY5jNapscN6G3wC0aNOpX3GMtOsWhImpXFf3SOUx+H18qfp3MLg=" } }, { "type": "step", "result": "=\\int_{0}^{\\frac{\\sqrt{2}}{2}}\\frac{1}{u}du" }, { "type": "step", "primary": "Use the common integral: $$\\int\\:\\frac{1}{u}du=\\ln\\left(\\left|u\\right|\\right)$$", "result": "=[\\ln\\left|u\\right|]_{0}^{\\frac{\\sqrt{2}}{2}}" }, { "type": "interim", "title": "$$\\frac{\\sqrt{2}}{2}=\\frac{1}{\\sqrt{2}}$$", "input": "\\frac{\\sqrt{2}}{2}", "steps": [ { "type": "step", "primary": "Apply radical rule: $$\\sqrt[n]{a}=a^{\\frac{1}{n}}$$", "secondary": [ "$$\\sqrt{2}=2^{\\frac{1}{2}}$$" ], "result": "=\\frac{2^{\\frac{1}{2}}}{2}", "meta": { "practiceLink": "/practice/radicals-practice", "practiceTopic": "Radical Rules" } }, { "type": "step", "primary": "Apply exponent rule: $$\\frac{x^{a}}{x^{b}}=\\frac{1}{x^{b-a}}$$", "secondary": [ "$$\\frac{2^{\\frac{1}{2}}}{2^{1}}=\\frac{1}{2^{1-\\frac{1}{2}}}$$" ], "result": "=\\frac{1}{2^{1-\\frac{1}{2}}}", "meta": { "practiceLink": "/practice/exponent-practice", "practiceTopic": "Expand FOIL" } }, { "type": "step", "primary": "Subtract the numbers: $$1-\\frac{1}{2}=\\frac{1}{2}$$", "result": "=\\frac{1}{2^{\\frac{1}{2}}}" }, { "type": "step", "primary": "Apply radical rule: $$a^{\\frac{1}{n}}=\\sqrt[n]{a}$$", "secondary": [ "$$2^{\\frac{1}{2}}=\\sqrt{2}$$" ], "result": "=\\frac{1}{\\sqrt{2}}", "meta": { "practiceLink": "/practice/radicals-practice", "practiceTopic": "Radical Rules" } } ], "meta": { "solvingClass": "Solver", "interimType": "Solver", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s74DwjiBDVBIb8+yh6fkkMMTc5Hq2e1dPlcHt0xrTeJGqrju+5Z51e/ZZSD3gRHwjBuvxaX7NiCEnOYksp0qBulxTvQPornS61GDWTUYpUdOMjigtccxNWJk7pEnO8OW1Uh+4bFcsDaYMbs7uBxyxhsu0lRFXlZWqEEXIzmPcPR+WwiNrEngO+NNvZ9sqNu+2V" } }, { "type": "step", "result": "=[\\ln\\left|u\\right|]_{0}^{\\frac{1}{\\sqrt{2}}}" }, { "type": "interim", "title": "Compute the boundaries:$${\\quad}$$diverges", "input": "[\\ln\\left|u\\right|]_{0}^{\\frac{1}{\\sqrt{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": "$$\\lim_{u\\to\\:0+}\\left(\\ln\\left|u\\right|\\right)=-\\infty\\:$$", "input": "\\lim_{u\\to\\:0+}\\left(\\ln\\left|u\\right|\\right)", "steps": [ { "type": "step", "primary": "$$u$$ is positive when $$u\\to\\:0+$$. Therefore $$\\left|u\\right|=u$$", "result": "=\\lim_{u\\to\\:0+}\\left(\\ln\\left(u\\right)\\right)" }, { "type": "step", "primary": "Apply the common limit: $$\\lim_{u\\to\\:0+}\\left(\\ln\\left(u\\right)\\right)=-\\infty\\:$$", "result": "=-\\infty\\:" } ], "meta": { "solvingClass": "Limits", "interimType": "Limits" } }, { "type": "interim", "title": "$$\\lim_{u\\to\\:\\frac{1}{\\sqrt{2}}-}\\left(\\ln\\left|u\\right|\\right)=-\\frac{1}{2}\\ln\\left(2\\right)$$", "input": "\\lim_{u\\to\\:\\frac{1}{\\sqrt{2}}-}\\left(\\ln\\left|u\\right|\\right)", "steps": [ { "type": "step", "primary": "Plug in the value $$u=\\frac{1}{\\sqrt{2}}$$", "result": "=\\ln\\left|\\frac{1}{\\sqrt{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": "interim", "title": "Simplify $$\\ln\\left|\\frac{1}{\\sqrt{2}}\\right|:{\\quad}-\\frac{1}{2}\\ln\\left(2\\right)$$", "input": "\\ln\\left|\\frac{1}{\\sqrt{2}}\\right|", "result": "=-\\frac{1}{2}\\ln\\left(2\\right)", "steps": [ { "type": "interim", "title": "Simplify $$\\left|\\frac{1}{\\sqrt{2}}\\right|:{\\quad}\\frac{1}{\\sqrt{2}}$$", "input": "\\left|\\frac{1}{\\sqrt{2}}\\right|", "result": "=\\ln\\left(\\frac{1}{\\sqrt{2}}\\right)", "steps": [ { "type": "step", "primary": "Apply absolute rule: $$\\left|a\\right|=a,\\:a\\ge0$$", "secondary": [ "$$\\left|\\frac{1}{\\sqrt{2}}\\right|=\\frac{1}{\\sqrt{2}}$$" ], "result": "=\\frac{1}{\\sqrt{2}}" } ], "meta": { "interimType": "Algebraic Manipulation Simplify Title 1Eq" } }, { "type": "step", "primary": "Apply log rule: $$\\log_{a}\\left(\\frac{1}{x}\\right)=-\\log_{a}\\left(x\\right)$$", "result": "=-\\ln\\left(\\sqrt{2}\\right)", "meta": { "practiceLink": "/practice/logarithms-practice", "practiceTopic": "Expand FOIL" } }, { "type": "step", "primary": "Rewrite as", "result": "=-\\ln\\left(2^{\\frac{1}{2}}\\right)" }, { "type": "step", "primary": "Apply log rule $$\\log_{a}\\left(x^b\\right)=b\\cdot\\log_{a}\\left(x\\right),\\:\\quad$$ assuming $$x\\:\\geq\\:0$$", "result": "=-\\frac{1}{2}\\ln\\left(2\\right)" } ], "meta": { "solvingClass": "Solver", "interimType": "Generic Simplify Specific 1Eq", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7uDlvWcMTI+CAEcrqD2avcD1HpTtaoZnV1XBcXFAQZLVozLZZSK2jC/Pd8a5g5tESA585Wz2Y8ioMtXlAhbC3efr9B9ajo0t7o4LpgosShLzS4MDEoHe9+4vYgu9wEvJ38LfSxJ+0AgVLpCSnLX0iSrFIab4jt/vcKLUwgOzi/lw10K1IsG2UtLzH2W9zXgHtv+nudFmpXG7F1NaL2WbkI3XcveFiu0trlVqTKOfHoNQ=" } } ], "meta": { "solvingClass": "Limits", "interimType": "Limits" } }, { "type": "step", "result": "=\\mathrm{diverges}" }, { "type": "step", "primary": "Since $$\\lim_{u\\to\\:0+}\\left(\\ln\\left|u\\right|\\right)=-\\infty\\:$$", "result": "=\\mathrm{diverges}" } ], "meta": { "interimType": "Integral Definite Limit Boundaries 0Eq", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7l7+Hp9GxvB/5I+4/5L7s7/FfqYwNzMzIMwjVA5a1PadPrZr1YEt17AAaJ6hxcb2WZ81F+WiYxzp3QrzjnLPYCFcwrjcVB4eQQfx4q+i/BIu6ace++vzlSjofnb5F9AU9YDI1lJ+LoqdA3z0Fd4M7qR1b3CgbC0GwjRsL4v1uEimBBTEk/JQ2cZ9WKuRzClU7gX3LUoUh4eqNSlXiuujzcFZ+pBVREKCh2SNPS3M2FOM=" } }, { "type": "step", "result": "=\\mathrm{diverges}" } ], "meta": { "solvingClass": "Integrals" } }, "meta": { "showVerify": true } }