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Definite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals5 Marks
Question
Evaluate the following integrals:$\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{5+4\sin\text{x}}\text{ dx}$
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Answer

Let $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{5+4\sin\text{x}}\text{ dx}$ Then,$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{5+4\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{1+\tan^2\frac{\text{x}}{2}}{5\Big(1+\tan^2\frac{\text{x}}{2}\Big)+8\tan\frac{\text{x}}{2}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\sec^2\frac{\text{x}}{2}}{5\tan^2\frac{\text{x}}{2}+8\tan\frac{\text{x}}{2}+5}\text{ dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$ Then, $\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{ dx}=\text{dt}$ When $\text{x}=0,\text{t}=0$ and $\text{x}=\frac{\pi}{2},\text{t}=1$$\therefore\ \text{I}=2\int\limits_0^1\frac{1}{5\text{t}^2+8\text{t}+5}\text{ dt}$
$\Rightarrow\text{I}=2\int\limits_0^1\frac{1}{\big(\sqrt{5\text{t}}\big)^2+8\text{t}+5+\Big(\frac{4}{\sqrt{5}}\Big)^2-\Big(\frac{4}{\sqrt{5}}\Big)^2}\text{ dt}$
$\Rightarrow\text{I}=2\int\limits_0^1\frac{1}{\Big(\sqrt{5}\text{t}+\frac{4}{\sqrt{5}}\Big)^2+\frac{9}{5}}\text{ dt}$
$\Rightarrow\text{I}=\frac{2}{3}\begin{bmatrix}\tan^{-1}\begin{pmatrix}\frac{\sqrt{5}\text{t}+\frac{4}{\sqrt{5}}}{\frac{3}{\sqrt{5}}} \end{pmatrix}\end{bmatrix}^1_0$
$\Rightarrow\text{I}=\frac{2}{3}\Big[\tan^{-1}3-\tan^{-1}\frac{4}{3}\Big]$
$\Rightarrow\text{I}=\frac{2}{3}\Bigg[\tan^{-1}\Bigg(\frac{3-\frac{4}{3}}{1+3\times\frac{4}{3}}\Bigg)\Bigg]$
$\Rightarrow\text{I}=\frac{2}{3}\tan^{-1}\frac{1}{3}$

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