# 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.