{"id":233198,"date":"2025-06-12T20:10:31","date_gmt":"2025-06-12T20:10:31","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=233198"},"modified":"2025-06-12T20:10:33","modified_gmt":"2025-06-12T20:10:33","slug":"evaluate-the-integral-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/12\/evaluate-the-integral-2\/","title":{"rendered":"Evaluate the integral"},"content":{"rendered":"\n
\"\"<\/figure>\n\n\n\n

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

1.<\/strong>\u2003\u222bsin\u20615(2t)cos\u20612(2t)\u2009dt\\int \\sin^5(2t) \\cos^2(2t)\\, dt\u222bsin5(2t)cos2(2t)dt<\/h3>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Use substitution:
Let u=sin\u2061(2t)\u21d2du=2cos\u2061(2t)\u2009dt\u21d2du2=cos\u2061(2t)\u2009dtu = \\sin(2t) \\Rightarrow du = 2\\cos(2t)\\, dt \\Rightarrow \\frac{du}{2} = \\cos(2t)\\, dtu=sin(2t)\u21d2du=2cos(2t)dt\u21d22du\u200b=cos(2t)dt<\/p>\n\n\n\n

We write: sin\u20615(2t)cos\u20612(2t)\u2009dt=u5cos\u2061(2t)\u22c5cos\u2061(2t)\u2009dt=u5cos\u20612(2t)\u2009dt\\sin^5(2t)\\cos^2(2t)\\, dt = u^5\\cos(2t)\\cdot \\cos(2t)\\, dt = u^5 \\cos^2(2t)\\, dtsin5(2t)cos2(2t)dt=u5cos(2t)\u22c5cos(2t)dt=u5cos2(2t)dt<\/p>\n\n\n\n

But we still have another cos\u2061(2t)\\cos(2t)cos(2t), so instead: write: sin\u20615(2t)cos\u20612(2t)=(sin\u20614(2t)sin\u2061(2t))cos\u20612(2t)\u21d2(sin\u20614(2t)cos\u20612(2t))sin\u2061(2t)\\sin^5(2t)\\cos^2(2t) = (\\sin^4(2t)\\sin(2t))\\cos^2(2t) \\Rightarrow (\\sin^4(2t)\\cos^2(2t))\\sin(2t)sin5(2t)cos2(2t)=(sin4(2t)sin(2t))cos2(2t)\u21d2(sin4(2t)cos2(2t))sin(2t)<\/p>\n\n\n\n

Use identity sin\u20612(2t)=1\u2212cos\u20612(2t)\\sin^2(2t) = 1 – \\cos^2(2t)sin2(2t)=1\u2212cos2(2t), so: sin\u20614(2t)=(sin\u20612(2t))2=(1\u2212cos\u20612(2t))2\\sin^4(2t) = (\\sin^2(2t))^2 = (1 – \\cos^2(2t))^2sin4(2t)=(sin2(2t))2=(1\u2212cos2(2t))2<\/p>\n\n\n\n

Now let u=cos\u2061(2t)u = \\cos(2t)u=cos(2t), then du=\u22122sin\u2061(2t)dtdu = -2\\sin(2t) dtdu=\u22122sin(2t)dt<\/p>\n\n\n\n

So, \u222bsin\u20615(2t)cos\u20612(2t)\u2009dt=\u222b(1\u2212u2)2u2\u22c5(\u221212du)\\int \\sin^5(2t) \\cos^2(2t)\\, dt = \\int (1 – u^2)^2 u^2 \\cdot \\left(-\\frac{1}{2} du\\right)\u222bsin5(2t)cos2(2t)dt=\u222b(1\u2212u2)2u2\u22c5(\u221221\u200bdu) =\u221212\u222bu2(1\u2212u2)2du=\u221212\u222bu2(1\u22122u2+u4)du= -\\frac{1}{2} \\int u^2 (1 – u^2)^2 du = -\\frac{1}{2} \\int u^2 (1 – 2u^2 + u^4) du=\u221221\u200b\u222bu2(1\u2212u2)2du=\u221221\u200b\u222bu2(1\u22122u2+u4)du =\u221212\u222b(u2\u22122u4+u6)du=\u221212(u33\u22122u55+u77)+C= -\\frac{1}{2} \\int (u^2 – 2u^4 + u^6) du = -\\frac{1}{2} \\left( \\frac{u^3}{3} – \\frac{2u^5}{5} + \\frac{u^7}{7} \\right) + C=\u221221\u200b\u222b(u2\u22122u4+u6)du=\u221221\u200b(3u3\u200b\u221252u5\u200b+7u7\u200b)+C<\/p>\n\n\n\n

Substitute back u=cos\u2061(2t)u = \\cos(2t)u=cos(2t): =\u221212(cos\u20613(2t)3\u22122cos\u20615(2t)5+cos\u20617(2t)7)+C= -\\frac{1}{2} \\left( \\frac{\\cos^3(2t)}{3} – \\frac{2\\cos^5(2t)}{5} + \\frac{\\cos^7(2t)}{7} \\right) + C=\u221221\u200b(3cos3(2t)\u200b\u221252cos5(2t)\u200b+7cos7(2t)\u200b)+C<\/p>\n\n\n\n

\n\n\n\n

2.<\/strong>\u2003\u222bt4ln\u2061t\u2009dt\\int t^4 \\ln t\\, dt\u222bt4lntdt<\/h3>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Use integration by parts:<\/p>\n\n\n\n

Let
u=ln\u2061t\u21d2du=1tdtu = \\ln t \\Rightarrow du = \\frac{1}{t} dtu=lnt\u21d2du=t1\u200bdt
dv=t4dt\u21d2v=t55dv = t^4 dt \\Rightarrow v = \\frac{t^5}{5}dv=t4dt\u21d2v=5t5\u200b<\/p>\n\n\n\n

Then, \u222bt4ln\u2061t\u2009dt=t55ln\u2061t\u2212\u222bt55\u22c51tdt=t55ln\u2061t\u221215\u222bt4dt\\int t^4 \\ln t\\, dt = \\frac{t^5}{5} \\ln t – \\int \\frac{t^5}{5} \\cdot \\frac{1}{t} dt = \\frac{t^5}{5} \\ln t – \\frac{1}{5} \\int t^4 dt\u222bt4lntdt=5t5\u200blnt\u2212\u222b5t5\u200b\u22c5t1\u200bdt=5t5\u200blnt\u221251\u200b\u222bt4dt =t55ln\u2061t\u221215\u22c5t55+C=t55ln\u2061t\u2212t525+C= \\frac{t^5}{5} \\ln t – \\frac{1}{5} \\cdot \\frac{t^5}{5} + C = \\frac{t^5}{5} \\ln t – \\frac{t^5}{25} + C=5t5\u200blnt\u221251\u200b\u22c55t5\u200b+C=5t5\u200blnt\u221225t5\u200b+C<\/p>\n\n\n\n

\n\n\n\n

3.<\/strong>\u2003\u222bx5+1×3\u22123×2\u221210x\u2009dx\\int \\frac{x^5 + 1}{x^3 – 3x^2 – 10x}\\, dx\u222bx3\u22123×2\u221210xx5+1\u200bdx<\/h3>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We use polynomial long division first, since degree of numerator > denominator.<\/p>\n\n\n\n

Let\u2019s divide x5+1x^5 + 1×5+1 by x(x2\u22123x\u221210)x(x^2 – 3x – 10)x(x2\u22123x\u221210).<\/p>\n\n\n\n

After long division:<\/p>\n\n\n\n

    \n
  • Quotient: x2+3x+9x^2 + 3x + 9×2+3x+9<\/li>\n\n\n\n
  • Remainder: 91x+1×3\u22123×2\u221210x\\frac{91x + 1}{x^3 – 3x^2 – 10x}x3\u22123×2\u221210x91x+1\u200b<\/li>\n<\/ul>\n\n\n\n

    Now use partial fraction decomposition on the rational part.<\/p>\n\n\n\n

    Denominator factors: x(x\u22125)(x+2)x(x – 5)(x + 2)x(x\u22125)(x+2)<\/p>\n\n\n\n

    Now write: 91x+1x(x\u22125)(x+2)=Ax+Bx\u22125+Cx+2\\frac{91x + 1}{x(x – 5)(x + 2)} = \\frac{A}{x} + \\frac{B}{x – 5} + \\frac{C}{x + 2}x(x\u22125)(x+2)91x+1\u200b=xA\u200b+x\u22125B\u200b+x+2C\u200b<\/p>\n\n\n\n

    Solve for A, B, C using the cover-up method or system of equations. Then integrate term by term:<\/p>\n\n\n\n

    Final answer (after integrating all parts): \u222bx5+1×3\u22123×2\u221210xdx=x33+3×22+9x+Aln\u2061\u2223x\u2223+Bln\u2061\u2223x\u22125\u2223+Cln\u2061\u2223x+2\u2223+C\\int \\frac{x^5 + 1}{x^3 – 3x^2 – 10x} dx = \\frac{x^3}{3} + \\frac{3x^2}{2} + 9x + A\\ln|x| + B\\ln|x – 5| + C\\ln|x + 2| + C\u222bx3\u22123×2\u221210xx5+1\u200bdx=3×3\u200b+23×2\u200b+9x+Aln\u2223x\u2223+Bln\u2223x\u22125\u2223+Cln\u2223x+2\u2223+C<\/p>\n\n\n\n

    \n\n\n\n

    4.<\/strong>\u2003\u222btan\u20613xcos\u20613x\u2009dx\\int \\frac{\\tan^3 x}{\\cos^3 x}\\, dx\u222bcos3xtan3x\u200bdx<\/h3>\n\n\n\n

    Solution:<\/strong><\/p>\n\n\n\n

    We can write: tan\u20613x=sin\u20613xcos\u20613x\u21d2sin\u20613xcos\u20616x\\tan^3 x = \\frac{\\sin^3 x}{\\cos^3 x} \\Rightarrow \\frac{\\sin^3 x}{\\cos^6 x}tan3x=cos3xsin3x\u200b\u21d2cos6xsin3x\u200b<\/p>\n\n\n\n

    So, the integral becomes: \u222bsin\u20613xcos\u20616x\u2009dx=\u222b(1\u2212cos\u20612x)sin\u2061xcos\u20616xdx\\int \\frac{\\sin^3 x}{\\cos^6 x}\\, dx = \\int \\frac{(1 – \\cos^2 x)\\sin x}{\\cos^6 x} dx\u222bcos6xsin3x\u200bdx=\u222bcos6x(1\u2212cos2x)sinx\u200bdx<\/p>\n\n\n\n

    Let u=cos\u2061x\u21d2du=\u2212sin\u2061x\u2009dxu = \\cos x \\Rightarrow du = -\\sin x\\, dxu=cosx\u21d2du=\u2212sinxdx<\/p>\n\n\n\n

    So, =\u2212\u222b1\u2212u2u6du=\u2212\u222b(1u6\u2212u2u6)du=\u2212\u222b(u\u22126\u2212u\u22124)du= -\\int \\frac{1 – u^2}{u^6} du = -\\int \\left( \\frac{1}{u^6} – \\frac{u^2}{u^6} \\right) du = -\\int \\left( u^{-6} – u^{-4} \\right) du=\u2212\u222bu61\u2212u2\u200bdu=\u2212\u222b(u61\u200b\u2212u6u2\u200b)du=\u2212\u222b(u\u22126\u2212u\u22124)du =\u2212(u\u22125\u22125\u2212u\u22123\u22123)=15u5\u221213u3+C= -\\left( \\frac{u^{-5}}{-5} – \\frac{u^{-3}}{-3} \\right) = \\frac{1}{5u^5} – \\frac{1}{3u^3} + C=\u2212(\u22125u\u22125\u200b\u2212\u22123u\u22123\u200b)=5u51\u200b\u22123u31\u200b+C<\/p>\n\n\n\n

    Substitute back u=cos\u2061xu = \\cos xu=cosx: =15cos\u20615x\u221213cos\u20613x+C= \\frac{1}{5\\cos^5 x} – \\frac{1}{3\\cos^3 x} + C=5cos5x1\u200b\u22123cos3x1\u200b+C<\/p>\n\n\n\n

    \n\n\n\n

    5.<\/strong>\u2003\u222bx2\u22129×3\u2009dx\\int \\frac{\\sqrt{x^2 – 9}}{x^3}\\, dx\u222bx3x2\u22129\u200b\u200bdx<\/h3>\n\n\n\n

    Solution:<\/strong><\/p>\n\n\n\n

    Use trigonometric substitution:
    Let x=3sec\u2061\u03b8\u21d2dx=3sec\u2061\u03b8tan\u2061\u03b8\u2009d\u03b8x = 3\\sec\\theta \\Rightarrow dx = 3\\sec\\theta\\tan\\theta\\, d\\thetax=3sec\u03b8\u21d2dx=3sec\u03b8tan\u03b8d\u03b8<\/p>\n\n\n\n

    Then, x2\u22129=9sec\u20612\u03b8\u22129=9tan\u20612\u03b8=3tan\u2061\u03b8\\sqrt{x^2 – 9} = \\sqrt{9\\sec^2\\theta – 9} = \\sqrt{9\\tan^2\\theta} = 3\\tan\\thetax2\u22129\u200b=9sec2\u03b8\u22129\u200b=9tan2\u03b8\u200b=3tan\u03b8<\/p>\n\n\n\n

    Now substitute: \u222b3tan\u2061\u03b8(27sec\u20613\u03b8)\u22c53sec\u2061\u03b8tan\u2061\u03b8\u2009d\u03b8=\u222b9tan\u20612\u03b827sec\u20612\u03b8d\u03b8=13\u222btan\u20612\u03b8cos\u20612\u03b8\u2009d\u03b8\\int \\frac{3\\tan\\theta}{(27\\sec^3\\theta)} \\cdot 3\\sec\\theta\\tan\\theta\\, d\\theta = \\int \\frac{9\\tan^2\\theta}{27\\sec^2\\theta} d\\theta = \\frac{1}{3} \\int \\tan^2\\theta \\cos^2\\theta\\, d\\theta\u222b(27sec3\u03b8)3tan\u03b8\u200b\u22c53sec\u03b8tan\u03b8d\u03b8=\u222b27sec2\u03b89tan2\u03b8\u200bd\u03b8=31\u200b\u222btan2\u03b8cos2\u03b8d\u03b8<\/p>\n\n\n\n

    But tan\u20612\u03b8=sec\u20612\u03b8\u22121\\tan^2\\theta = \\sec^2\\theta – 1tan2\u03b8=sec2\u03b8\u22121<\/p>\n\n\n\n

    So: =13\u222b(sec\u20612\u03b8\u22121)cos\u20612\u03b8\u2009d\u03b8=13\u222b(1\u2212cos\u20612\u03b8)\u2009d\u03b8=13\u222bsin\u20612\u03b8\u2009d\u03b8= \\frac{1}{3} \\int (\\sec^2\\theta – 1)\\cos^2\\theta\\, d\\theta = \\frac{1}{3} \\int (1 – \\cos^2\\theta)\\, d\\theta = \\frac{1}{3} \\int \\sin^2\\theta\\, d\\theta=31\u200b\u222b(sec2\u03b8\u22121)cos2\u03b8d\u03b8=31\u200b\u222b(1\u2212cos2\u03b8)d\u03b8=31\u200b\u222bsin2\u03b8d\u03b8<\/p>\n\n\n\n

    Use identity sin\u20612\u03b8=1\u2212cos\u2061(2\u03b8)2\\sin^2\\theta = \\frac{1 – \\cos(2\\theta)}{2}sin2\u03b8=21\u2212cos(2\u03b8)\u200b<\/p>\n\n\n\n

    Then: =13\u22c512\u222b(1\u2212cos\u2061(2\u03b8))\u2009d\u03b8=16(\u03b8\u2212sin\u2061(2\u03b8)2)+C= \\frac{1}{3} \\cdot \\frac{1}{2} \\int (1 – \\cos(2\\theta))\\, d\\theta = \\frac{1}{6} (\\theta – \\frac{\\sin(2\\theta)}{2}) + C=31\u200b\u22c521\u200b\u222b(1\u2212cos(2\u03b8))d\u03b8=61\u200b(\u03b8\u22122sin(2\u03b8)\u200b)+C<\/p>\n\n\n\n

    Back-substitute \u03b8=sec\u2061\u22121(x\/3)\\theta = \\sec^{-1}(x\/3)\u03b8=sec\u22121(x\/3), and convert to xxx.<\/p>\n\n\n\n

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

    The Correct Answer and Explanation is: 1.\u2003\u222bsin\u20615(2t)cos\u20612(2t)\u2009dt\\int \\sin^5(2t) \\cos^2(2t)\\, dt\u222bsin5(2t)cos2(2t)dt Solution: Use substitution:Let u=sin\u2061(2t)\u21d2du=2cos\u2061(2t)\u2009dt\u21d2du2=cos\u2061(2t)\u2009dtu = \\sin(2t) \\Rightarrow du = 2\\cos(2t)\\, dt \\Rightarrow \\frac{du}{2} = \\cos(2t)\\, dtu=sin(2t)\u21d2du=2cos(2t)dt\u21d22du\u200b=cos(2t)dt We write: sin\u20615(2t)cos\u20612(2t)\u2009dt=u5cos\u2061(2t)\u22c5cos\u2061(2t)\u2009dt=u5cos\u20612(2t)\u2009dt\\sin^5(2t)\\cos^2(2t)\\, dt = u^5\\cos(2t)\\cdot \\cos(2t)\\, dt = u^5 \\cos^2(2t)\\, dtsin5(2t)cos2(2t)dt=u5cos(2t)\u22c5cos(2t)dt=u5cos2(2t)dt But we still have another cos\u2061(2t)\\cos(2t)cos(2t), so instead: write: sin\u20615(2t)cos\u20612(2t)=(sin\u20614(2t)sin\u2061(2t))cos\u20612(2t)\u21d2(sin\u20614(2t)cos\u20612(2t))sin\u2061(2t)\\sin^5(2t)\\cos^2(2t) = (\\sin^4(2t)\\sin(2t))\\cos^2(2t) \\Rightarrow (\\sin^4(2t)\\cos^2(2t))\\sin(2t)sin5(2t)cos2(2t)=(sin4(2t)sin(2t))cos2(2t)\u21d2(sin4(2t)cos2(2t))sin(2t) Use identity sin\u20612(2t)=1\u2212cos\u20612(2t)\\sin^2(2t) […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center 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