Question

Evaluate the integral: $\int_{0}^{\frac{π}{2}} \frac{1}{5 + 4 \sin x} d x$ .

Accepted Answer

Let $I = \int_{0}^{\frac{π}{2}} \frac{1}{5 + 4 \sin x} d x$

Substitute $\sin x = \frac{2 \tan (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})}$

Therefore, $I = \int_{0}^{\frac{π}{2}} \frac{1}{5 + 4 \frac{2 \tan (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})}} d x$

$\Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{1 + \tan^{2} (\frac{x}{2})}{5 (1 + \tan^{2} (\frac{x}{2})) + 8 \tan (\frac{x}{2})} d x$

$\Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\sec^{2} (\frac{x}{2})}{5 + 5 \tan^{2} (\frac{x}{2}) + 8 \tan (\frac{x}{2})} d x$

Put $\tan (\frac{x}{2}) = t$

Differentiate w.r.t $x$ on both sides, we get: $\frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t$

Also, $[x = 0 \Rightarrow t = 0]$ and $[x = \frac{π}{2} \Rightarrow t = 1]$

Therefore, $I = \int_{0}^{1} \frac{2}{5 + 5 t^{2} + 8 t} d t$

$\Rightarrow I = \frac{1}{5} \int_{0}^{1} \frac{2}{1 + t^{2} + \frac{8}{5} t} d t$

$\Rightarrow I = \frac{2}{5} \int_{0}^{1} \frac{1}{t^{2} + \frac{8}{5} t + \frac{16}{25} + \frac{9}{25}} d t$

$\Rightarrow I = \frac{2}{5} \int_{0}^{1} \frac{1}{{(t + \frac{4}{5})}^{2} + \frac{9}{25}} d t$

$\Rightarrow I = \frac{2}{5} \int_{0}^{1} \frac{1}{{(t + \frac{4}{5})}^{2} + {(\frac{3}{5})}^{2}} d t$

$\Rightarrow I = \frac{2}{5} \times \frac{5}{3} {[\tan^{- 1} (\frac{t + \frac{4}{5}}{\frac{3}{5}})]}_{0}^{1} [∵ \int \frac{1}{a^{2} + x^{2}} d x = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + C]$

$\Rightarrow I = \frac{2}{5} \times \frac{5}{3} {[\tan^{- 1} (\frac{5 t + 4}{3})]}_{0}^{1}$

$\Rightarrow I = \frac{2}{3} (\tan^{- 1} 3 - \tan^{- 1} (\frac{4}{3}))$

$\Rightarrow I = \frac{2}{3} \times \tan^{- 1} (\frac{3 - \frac{4}{3}}{5})$ . $\Rightarrow I = \frac{2}{3} \tan^{- 1} (\frac{1}{3}) [∵ \tan^{- 1} x - \tan^{- 1} y = \tan^{- 1} (\frac{x - y}{1 + x y}), x y > 1]$

Hence, $\int_{0}^{\frac{π}{2}} \frac{1}{5 + 4 \sin x} d x = \frac{2}{3} \tan^{- 1} (\frac{1}{3})$ .

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