Integration Examples

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

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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

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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.