Example 1 : <\/span>Evaluate : \\(\\int\\) \\(dx\\over {3sinx + 4cosx}\\)<\/p>\n Solution : <\/span>I = \\(\\int\\) \\(dx\\over {3sinx + 4cosx}\\) = \\(\\int\\) \\(dx\\over {3[{2tan{x\\over 2}\\over {1+tan^2{x\\over 2}}}] + 4[{1-tan^2{x\\over 2}\\over {1+tan^2{x\\over 2}}}]}\\) = \\(\\int\\) \\(sec^2{x\\over 2}dx\\over {4+6tan{x\\over 2}-4tan^2{x\\over 2}}\\) Example 2 : <\/span> Evaluate : \\(\\int\\) \\(cos^4xdx\\over {sin^3x{(sin^5x + cos^5x)^{3\\over 5}}}\\)<\/p>\n Solution : <\/span>I = \\(\\int\\) \\(cos^4xdx\\over {sin^3x{(sin^5x + cos^5x)^{3\\over 5}}}\\) Example 3 : <\/span>Prove that \\(\\int_{0}^{\\pi\/2}\\) log(sinx)dx = \\(\\int_{0}^{\\pi\/2}\\) log(cosx)dx = -\\(\\pi\\over 2\\)log 2.<\/p>\n Solution : <\/span>Let I = \\(\\int_{0}^{\\pi\/2}\\) log(sinx)dx …(i) Practice these given integration examples to test your knowledge on concepts of integration.<\/p>\n \n <\/div>\n <\/div>\n","protected":false},"excerpt":{"rendered":" Here you will learn some integration examples for better understanding of integration concepts. Example 1 : Evaluate : \\(\\int\\) \\(dx\\over {3sinx + 4cosx}\\) Solution : I = \\(\\int\\) \\(dx\\over {3sinx + 4cosx}\\) = \\(\\int\\) \\(dx\\over {3[{2tan{x\\over 2}\\over {1+tan^2{x\\over 2}}}] + 4[{1-tan^2{x\\over 2}\\over {1+tan^2{x\\over 2}}}]}\\) = \\(\\int\\) \\(sec^2{x\\over 2}dx\\over {4+6tan{x\\over 2}-4tan^2{x\\over 2}}\\) let \\(tan{x\\over 2}\\) = …<\/p>\n
\n let \\(tan{x\\over 2}\\) = t, \\(\\therefore\\) \\({1\\over 2}sec^2{x\\over 2}\\)dx = dt
\n so I = \\(\\int\\) \\(2dt\\over {4+6t-4t^2}\\) = \\(1\\over 2\\) \\(\\int\\) \\(dt\\over {1-(t^2-{3\\over 2}t})\\) = \\(1\\over 2\\) \\(\\int\\) \\(dt\\over {{25\\over 16}-{(t-{3\\over 4})}^2}\\)
\n = \\(1\\over 2\\) \\(1\\over {2({5\\over 4})}\\) \\(ln|{{{5\\over 4}+(t-{3\\over 4})}\\over {{5\\over 4}-(t-{3\\over 4})}}|\\) + C = \\(1\\over 5\\) \\(ln|{1+2tan{x\\over 2}\\over {4-2tan{x\\over 2}}}|\\) + C<\/p>\n
\n\n
\n = \\(\\int\\) \\(cos^4xdx\\over {sin^6x{(1 + cot^5x)^{3\\over 5}}}\\) = \\(\\int\\) \\(cot^4xcosec^2xdx\\over {{(1 + cot^5x)^{3\\over 5}}}\\)
\n Put \\(1+cot^5x\\) = t
\n \\(5cot^4xcosec^2x\\)dx = -dt
\n = -\\(1\\over 5\\) \\(\\int\\) \\(dt\\over {t^{3\/5}}\\) = -\\(1\\over 2\\) \\(t^{2\/5}\\) + C
\n = -\\(1\\over 2\\) \\({(1+cot^5x)}^{2\/5}\\) + C\n <\/p>
\n\n \n
\n then I = \\(\\int_{0}^{\\pi\/2}\\) \\(log(sin({\\pi\\over 2}-x))\\)dx = \\(\\int_{0}^{\\pi\/2}\\) log(cosx)dx …(ii)
\n Adding (i) and (ii), we get
\n 2I = \\(\\int_{0}^{\\pi\/2}\\) log(sinx)dx + \\(\\int_{0}^{\\pi\/2}\\) log(cosx)dx = \\(\\int_{0}^{\\pi\/2}\\) (log(sinx)dx + log(cosx))dx
\n \\(\\implies\\) \\(\\int_{0}^{\\pi\/2}\\) log(sinxcosx)dx = \\(\\int_{0}^{\\pi\/2}\\) \\(log({2sinxcosx\\over 2})\\)dx
\n = \\(\\int_{0}^{\\pi\/2}\\) \\(log({sin2x\\over 2})\\)dx = \\(\\int_{0}^{\\pi\/2}\\) log(sin2x)dx – \\(\\int_{0}^{\\pi\/2}\\) log(2)dx
\n = \\(\\int_{0}^{\\pi\/2}\\) log(sin2x)dx – (log 2)\\({(x)}^{\\pi\/2}_{0}\\)
\n \\(\\implies\\) 2I = \\(\\int_{0}^{\\pi\/2}\\) log(sin2x)dx – \\(\\pi\\over 2\\)log 2 …(iii)
\n Let \\(I_1\\) = \\(\\int_{0}^{\\pi\/2}\\) log(sin2x)dx, putting 2x = t, we get
\n \\(I_1\\) = \\(\\int_{0}^{\\pi}\\) log(sint)\\(dt\\over 2\\) = \\(1\\over 2\\) \\(\\int_{0}^{\\pi}\\) log(sint)dt = \\(1\\over 2\\) 2\\(\\int_{0}^{\\pi\/2}\\) log(sint)dt
\n \\(I_1\\) = \\(\\int_{0}^{\\pi\/2}\\) log(sinx)dx
\n \\(\\therefore\\) (iii) becomes; 2I = I – \\(\\pi\\over 2\\)log 2
\n Hence \\(\\int_{0}^{\\pi\/2}\\) log(sinx)dx = – \\(\\pi\\over 2\\)log 2<\/p>\n
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