Integration Examples

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 \(tan{x\over 2}\) = t,     \(\therefore\)   \({1\over 2}sec^2{x\over 2}\)dx = dt

so I = \(\int\) \(2dt\over {4+6t-4t^2}\) = \(1\over 2\) \(\int\) \(dt\over {1-(t^2-{3\over 2}t})\) = \(1\over 2\) \(\int\) \(dt\over {{25\over 16}-{(t-{3\over 4})}^2}\)

= \(1\over 2\) \(1\over {2({5\over 4})}\) \(ln|{{{5\over 4}+(t-{3\over 4})}\over {{5\over 4}-(t-{3\over 4})}}|\) + C = \(1\over 5\) \(ln|{1+2tan{x\over 2}\over {4-2tan{x\over 2}}}|\) + C



Example 2 : Evaluate : \(\int\) \(cos^4xdx\over {sin^3x{(sin^5x + cos^5x)^{3\over 5}}}\)

Solution : I = \(\int\) \(cos^4xdx\over {sin^3x{(sin^5x + cos^5x)^{3\over 5}}}\)

= \(\int\) \(cos^4xdx\over {sin^6x{(1 + cot^5x)^{3\over 5}}}\) = \(\int\) \(cot^4xcosec^2xdx\over {{(1 + cot^5x)^{3\over 5}}}\)

Put \(1+cot^5x\) = t

\(5cot^4xcosec^2x\)dx = -dt

= -\(1\over 5\) \(\int\) \(dt\over {t^{3/5}}\) = -\(1\over 2\) \(t^{2/5}\) + C

= -\(1\over 2\) \({(1+cot^5x)}^{2/5}\) + C



Example 3 : Prove that \(\int_{0}^{\pi/2}\) log(sinx)dx = \(\int_{0}^{\pi/2}\) log(cosx)dx = -\(\pi\over 2\)log 2.

Solution : Let I = \(\int_{0}^{\pi/2}\) log(sinx)dx     …(i)

then I = \(\int_{0}^{\pi/2}\) \(log(sin({\pi\over 2}-x))\)dx = \(\int_{0}^{\pi/2}\) log(cosx)dx     …(ii)

Adding (i) and (ii), we get

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

\(\implies\)   \(\int_{0}^{\pi/2}\) log(sinxcosx)dx = \(\int_{0}^{\pi/2}\) \(log({2sinxcosx\over 2})\)dx

= \(\int_{0}^{\pi/2}\) \(log({sin2x\over 2})\)dx = \(\int_{0}^{\pi/2}\) log(sin2x)dx – \(\int_{0}^{\pi/2}\) log(2)dx

= \(\int_{0}^{\pi/2}\) log(sin2x)dx – (log 2)\({(x)}^{\pi/2}_{0}\)

\(\implies\)   2I = \(\int_{0}^{\pi/2}\) log(sin2x)dx – \(\pi\over 2\)log 2     …(iii)

Let \(I_1\) = \(\int_{0}^{\pi/2}\) log(sin2x)dx,   putting 2x = t, we get

\(I_1\) = \(\int_{0}^{\pi}\) 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

\(I_1\) = \(\int_{0}^{\pi/2}\) log(sinx)dx

\(\therefore\)   (iii) becomes; 2I = I – \(\pi\over 2\)log 2

Hence   \(\int_{0}^{\pi/2}\) log(sinx)dx = – \(\pi\over 2\)log 2


Practice these given integration examples to test your knowledge on concepts of integration.

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