Math Integration Solution<\/title>\n \n \n <script src=\"https:\/\/polyfill.io\"><\/script>\n <script id=\"MathJax-script\" async src=\"https:\/\/jsdelivr.net\"><\/script>\n <script>\n window.MathJax = {\n tex: {\n inlineMath: [['$', '$'], ['\\\$', '\\\$']],\n displayMath: [['$$', '$$'], ['\\\\[', '\\\\]']]\n }\n };\n <\/script>\n\n <style>\n body {\n font-family: Arial, sans-serif;\n line-height: 1.6;\n margin: 40px auto;\n max-width: 800px;\n padding: 0 20px;\n color: #333;\n }\n .problem {\n font-weight: bold;\n font-size: 1.1em;\n margin-bottom: 20px;\n }\n .solution {\n background-color: #f9f9f9;\n border-left: 4px solid #00769b;\n padding: 20px;\n border-radius: 4px;\n }\n .step-table {\n margin-top: 15px;\n width: 100%;\n }\n <\/style>\n<\/head>\n<body>\n\n <div class=\"problem\">\n 1. Find the integral $\\int \\frac{1}{\\sqrt{a^2-x^2}} \\, dx$ with respect to $x$ and hence evaluate $\\int \\frac{1}{\\sqrt{9-25x^2}} \\, dx$.\n <\/div>\n\n <div class=\"solution\">\n <strong>Solution:<\/strong>\n <p>Let $I = \\int \\frac{1}{\\sqrt{a^2-x^2}} \\, dx$<\/p>\n \n <p>Substitute $x = a \\sin\\theta$, which means $\\theta = \\sin^{-1}\\left(\\frac{x}{a}\\right)$<\/p>\n <p>Differentiating both sides gives: $dx = a \\cos\\theta \\, d\\theta$<\/p>\n\n <p>Substituting these into the integral:<\/p>\n \n $$\n \\begin{align*}\n I &= \\int \\frac{a \\cos\\theta}{\\sqrt{a^2 – (a \\sin\\theta)^2}} \\, d\\theta \\\\[10pt]\n I &= \\int \\frac{a \\cos\\theta}{\\sqrt{a^2 (1 – \\sin^2\\theta)}} \\, d\\theta \\\\[10pt]\n I &= \\int \\frac{a \\cos\\theta}{a \\sqrt{\\cos^2\\theta}} \\, d\\theta \\\\[10pt]\n I &= \\int 1 \\, d\\theta \\\\[10pt]\n I &= \\theta + c \\\\[10pt]\n \\text{Therefore, } \\int \\frac{1}{\\sqrt{a^2-x^2}} \\, dx &= \\sin^{-1}\\left(\\frac{x}{a}\\right) + c\n \\end{align*}\n $$\n\n <hr style=\"border: 0; border-top: 1px solid #ccc; margin: 20px 0;\">\n\n <strong>Hence, evaluating the second part:<\/strong>\n <p>We need to find $\\int \\frac{1}{\\sqrt{9-25x^2}} \\, dx$. First, factor out $25$ from the square root to match our standard formula form:<\/p>\n \n $$\n \\int \\frac{1}{\\sqrt{25 \\left(\\frac{9}{25} – x^2\\right)}} \\, dx = \\int \\frac{1}{5\\sqrt{\\left(\\frac{3}{5}\\right)^2 – x^2}} \\, dx\n $$\n\n <p>Take the constant $\\frac{1}{5}$ outside the integral:<\/p>\n \n $$\n = \\frac{1}{5} \\int \\frac{1}{\\sqrt{\\left(\\frac{3}{5}\\right)^2 – x^2}} \\, dx\n $$\n\n <p>Using the formula $\\int \\frac{1}{\\sqrt{a^2-x^2}} \\, dx = \\sin^{-1}\\left(\\frac{x}{a}\\right) + c$, where $a = \\frac{3}{5}$:<\/p>\n \n $$\n = \\frac{1}{5} \\sin^{-1}\\left(\\frac{x}{\\frac{3}{5}}\\right) + c = \\frac{1}{5} \\sin^{-1}\\left(\\frac{5x}{3}\\right) + c\n $$\n <\/div>\n\n<\/body>\n<\/html>\n","protected":false},"excerpt":{"rendered":"<p>Math Integration Solution 1. Find the integral $\\int \\frac{1}{\\sqrt{a^2-x^2}} \\, dx$ with respect to $x$ and hence evaluate $\\int \\frac{1}{\\sqrt{9-25x^2}} \\, dx$. Solution: Let $I = \\int \\frac{1}{\\sqrt{a^2-x^2}} \\, dx$ Substitute $x = a \\sin\\theta$, which means $\\theta = \\sin^{-1}\\left(\\frac{x}{a}\\right)$ Differentiating both sides gives: $dx = a \\cos\\theta \\, d\\theta$ Substituting these into the integral: $$ \\begin{align*} I &=…..<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_monsterinsights_skip_tracking":false,"footnotes":""},"categories":[1],"tags":[],"class_list":["post-153","post","type-post","status-publish","format-standard","hentry","category-class-11"],"_links":{"self":[{"href":"https:\/\/mathsqna.com\/index.php?rest_route=\/wp\/v2\/posts\/153","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mathsqna.com\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mathsqna.com\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mathsqna.com\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mathsqna.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=153"}],"version-history":[{"count":1,"href":"https:\/\/mathsqna.com\/index.php?rest_route=\/wp\/v2\/posts\/153\/revisions"}],"predecessor-version":[{"id":154,"href":"https:\/\/mathsqna.com\/index.php?rest_route=\/wp\/v2\/posts\/153\/revisions\/154"}],"wp:attachment":[{"href":"https:\/\/mathsqna.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=153"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathsqna.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=153"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathsqna.com\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=153"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}

{"id":153,"date":"2026-10-24T18:07:15","date_gmt":"2026-10-24T18:07:15","guid":{"rendered":"https:\/\/mathsqna.com\/?p=153"},"modified":"2026-10-24T18:07:16","modified_gmt":"2026-10-24T18:07:16","slug":"integrations","status":"publish","type":"post","link":"https:\/\/mathsqna.com\/?p=153","title":{"rendered":"Integrations"},"content":{"rendered":"\n\n\n\n \n \n Math Integration Solution<\/title>\n \n \n <script src=\"https:\/\/polyfill.io\"><\/script>\n <script id=\"MathJax-script\" async src=\"https:\/\/jsdelivr.net\"><\/script>\n <script>\n window.MathJax = {\n tex: {\n inlineMath: [['$', '$'], ['\\\$', '\\\$']],\n displayMath: [['$$', '$$'], ['\\\\[', '\\\\]']]\n }\n };\n <\/script>\n\n <style>\n body {\n font-family: Arial, sans-serif;\n line-height: 1.6;\n margin: 40px auto;\n max-width: 800px;\n padding: 0 20px;\n color: #333;\n }\n .problem {\n font-weight: bold;\n font-size: 1.1em;\n margin-bottom: 20px;\n }\n .solution {\n background-color: #f9f9f9;\n border-left: 4px solid #00769b;\n padding: 20px;\n border-radius: 4px;\n }\n .step-table {\n margin-top: 15px;\n width: 100%;\n }\n <\/style>\n<\/head>\n<body>\n\n <div class=\"problem\">\n 1. Find the integral $\\int \\frac{1}{\\sqrt{a^2-x^2}} \\, dx$ with respect to $x$ and hence evaluate $\\int \\frac{1}{\\sqrt{9-25x^2}} \\, dx$.\n <\/div>\n\n <div class=\"solution\">\n <strong>Solution:<\/strong>\n <p>Let $I = \\int \\frac{1}{\\sqrt{a^2-x^2}} \\, dx$<\/p>\n \n <p>Substitute $x = a \\sin\\theta$, which means $\\theta = \\sin^{-1}\\left(\\frac{x}{a}\\right)$<\/p>\n <p>Differentiating both sides gives: $dx = a \\cos\\theta \\, d\\theta$<\/p>\n\n <p>Substituting these into the integral:<\/p>\n \n $$\n \\begin{align*}\n I &= \\int \\frac{a \\cos\\theta}{\\sqrt{a^2 – (a \\sin\\theta)^2}} \\, d\\theta \\\\[10pt]\n I &= \\int \\frac{a \\cos\\theta}{\\sqrt{a^2 (1 – \\sin^2\\theta)}} \\, d\\theta \\\\[10pt]\n I &= \\int \\frac{a \\cos\\theta}{a \\sqrt{\\cos^2\\theta}} \\, d\\theta \\\\[10pt]\n I &= \\int 1 \\, d\\theta \\\\[10pt]\n I &= \\theta + c \\\\[10pt]\n \\text{Therefore, } \\int \\frac{1}{\\sqrt{a^2-x^2}} \\, dx &= \\sin^{-1}\\left(\\frac{x}{a}\\right) + c\n \\end{align*}\n $$\n\n <hr style=\"border: 0; border-top: 1px solid #ccc; margin: 20px 0;\">\n\n <strong>Hence, evaluating the second part:<\/strong>\n <p>We need to find $\\int \\frac{1}{\\sqrt{9-25x^2}} \\, dx$. First, factor out $25$ from the square root to match our standard formula form:<\/p>\n \n $$\n \\int \\frac{1}{\\sqrt{25 \\left(\\frac{9}{25} – x^2\\right)}} \\, dx = \\int \\frac{1}{5\\sqrt{\\left(\\frac{3}{5}\\right)^2 – x^2}} \\, dx\n $$\n\n <p>Take the constant $\\frac{1}{5}$ outside the integral:<\/p>\n \n $$\n = \\frac{1}{5} \\int \\frac{1}{\\sqrt{\\left(\\frac{3}{5}\\right)^2 – x^2}} \\, dx\n $$\n\n <p>Using the formula $\\int \\frac{1}{\\sqrt{a^2-x^2}} \\, dx = \\sin^{-1}\\left(\\frac{x}{a}\\right) + c$, where $a = \\frac{3}{5}$:<\/p>\n \n $$\n = \\frac{1}{5} \\sin^{-1}\\left(\\frac{x}{\\frac{3}{5}}\\right) + c = \\frac{1}{5} \\sin^{-1}\\left(\\frac{5x}{3}\\right) + c\n $$\n <\/div>\n\n<\/body>\n<\/html>\n","protected":false},"excerpt":{"rendered":"<p>Math Integration Solution 1. Find the integral $\\int \\frac{1}{\\sqrt{a^2-x^2}} \\, dx$ with respect to $x$ and hence evaluate $\\int \\frac{1}{\\sqrt{9-25x^2}} \\, dx$. Solution: Let $I = \\int \\frac{1}{\\sqrt{a^2-x^2}} \\, dx$ Substitute $x = a \\sin\\theta$, which means $\\theta = \\sin^{-1}\\left(\\frac{x}{a}\\right)$ Differentiating both sides gives: $dx = a \\cos\\theta \\, d\\theta$ Substituting these into the integral: $$ \\begin{align*} I &=…..<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_monsterinsights_skip_tracking":false,"footnotes":""},"categories":[1],"tags":[],"class_list":["post-153","post","type-post","status-publish","format-standard","hentry","category-class-11"],"_links":{"self":[{"href":"https:\/\/mathsqna.com\/index.php?rest_route=\/wp\/v2\/posts\/153","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mathsqna.com\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mathsqna.com\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mathsqna.com\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mathsqna.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=153"}],"version-history":[{"count":1,"href":"https:\/\/mathsqna.com\/index.php?rest_route=\/wp\/v2\/posts\/153\/revisions"}],"predecessor-version":[{"id":154,"href":"https:\/\/mathsqna.com\/index.php?rest_route=\/wp\/v2\/posts\/153\/revisions\/154"}],"wp:attachment":[{"href":"https:\/\/mathsqna.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=153"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathsqna.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=153"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathsqna.com\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=153"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}