{"id":35658,"date":"2025-06-24T14:51:24","date_gmt":"2025-06-24T14:51:24","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=35658"},"modified":"2025-06-24T14:51:26","modified_gmt":"2025-06-24T14:51:26","slug":"find-the-indefinite-integral","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/find-the-indefinite-integral\/","title":{"rendered":"Find the indefinite integral"},"content":{"rendered":"\n
\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

The correct answer is:<\/p>\n\n\n\n

a.<\/strong> (sin\u2061(2x)\/30)\u22c5(4\u22125sin\u20612(2x)+3sin\u20614(2x))+C(\\sin(2x)\/30) \\cdot (4 – 5\\sin^2(2x) + 3\\sin^4(2x)) + C<\/p>\n\n\n\n

Explanation:<\/h3>\n\n\n\n

To solve the integral<\/p>\n\n\n\n

\u222bcos\u20613(2x)\u22c5sin\u20612(2x)\u2009dx,\\int \\cos^3(2x) \\cdot \\sin^2(2x) \\, dx,<\/p>\n\n\n\n

we start by expressing the odd power of cosine separately as<\/p>\n\n\n\n

cos\u20612(2x)\u22c5cos\u2061(2x).\\cos^2(2x) \\cdot \\cos(2x).<\/p>\n\n\n\n

Using the Pythagorean identity, we replace cos\u20612(2x)\\cos^2(2x) with 1\u2212sin\u20612(2x)1 – \\sin^2(2x), and then apply the substitution:<\/p>\n\n\n\n

u=sin\u2061(2x),du=2cos\u2061(2x)dx\u21d212du=cos\u2061(2x)dx.u = \\sin(2x), \\quad du = 2\\cos(2x) dx \\Rightarrow \\frac{1}{2}du = \\cos(2x) dx.<\/p>\n\n\n\n

Substitute into the integral:<\/p>\n\n\n\n

\u222b(1\u2212u2)u2\u22c512du=12\u222b(u2\u2212u4)du.\\int \\left(1 – u^2\\right) u^2 \\cdot \\frac{1}{2} du = \\frac{1}{2} \\int (u^2 – u^4) du.<\/p>\n\n\n\n

Now integrate term by term:<\/p>\n\n\n\n

12(u33\u2212u55)+C.\\frac{1}{2} \\left( \\frac{u^3}{3} – \\frac{u^5}{5} \\right) + C.<\/p>\n\n\n\n

To combine under a common denominator:<\/p>\n\n\n\n

12(5u3\u22123u515)=130u(5u2\u22123u4)+C.\\frac{1}{2} \\left( \\frac{5u^3 – 3u^5}{15} \\right) = \\frac{1}{30} u (5u^2 – 3u^4) + C.<\/p>\n\n\n\n

Now substitute back u=sin\u2061(2x)u = \\sin(2x):<\/p>\n\n\n\n

sin\u2061(2x)30(5sin\u20612(2x)\u22123sin\u20614(2x))+C.\\frac{\\sin(2x)}{30} \\left(5\\sin^2(2x) – 3\\sin^4(2x)\\right) + C.<\/p>\n\n\n\n

Factor out as:<\/p>\n\n\n\n

sin\u2061(2x)30(4\u22125sin\u20612(2x)+3sin\u20614(2x))+C.\\frac{\\sin(2x)}{30} \\left(4 – 5\\sin^2(2x) + 3\\sin^4(2x)\\right) + C.<\/p>\n\n\n\n

This matches choice a<\/strong>, so it is the correct answer. This technique uses trigonometric substitution to simplify the integrand, a standard approach when handling powers of sine and cosine.<\/p>\n\n\n\n

\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

The Correct Answer and Explanation is: The correct answer is: a. (sin\u2061(2x)\/30)\u22c5(4\u22125sin\u20612(2x)+3sin\u20614(2x))+C(\\sin(2x)\/30) \\cdot (4 – 5\\sin^2(2x) + 3\\sin^4(2x)) + C Explanation: To solve the integral \u222bcos\u20613(2x)\u22c5sin\u20612(2x)\u2009dx,\\int \\cos^3(2x) \\cdot \\sin^2(2x) \\, dx, we start by expressing the odd power of cosine separately as cos\u20612(2x)\u22c5cos\u2061(2x).\\cos^2(2x) \\cdot \\cos(2x). Using the Pythagorean identity, we replace cos\u20612(2x)\\cos^2(2x) with 1\u2212sin\u20612(2x)1 – […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-35658","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35658","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=35658"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35658\/revisions"}],"predecessor-version":[{"id":35661,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35658\/revisions\/35661"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=35658"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=35658"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=35658"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}