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Definite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals5 Marks
Question
Evaluate the following integrals:
$\int\limits_{0}^{\frac{\pi}{2}}\frac{\text{dx}}{\text{a}\cos\text{x}+\text{b}\sin\text{x}}\text{ a},\text{b}>0$
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Answer

$\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{\text{a}\cos\text{x}+\text{b}\sin\text{x}}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{\text{a}\Bigg(\frac{1-\tan^{2}\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}\Bigg)+\text{b}\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}\Bigg)}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\frac{\big(1+\tan^{2}\frac{\text{x}}{2}\big)}{\text{a}-\text{a}\tan^{2}\frac{\text{x}}{2}+2\text{b}\tan\frac{\text{x}}{2}}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\frac{\sec^2\frac{\text{x}}{2}}{\text{a}-\text{a}\tan^{2}\frac{\text{x}}{2}+2\text{b}\tan\frac{\text{x}}{2}}\text{ dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$ Then, $\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{ dx}=\text{dt}$
When $\text{x}=0,\text{t}=0,\text{x}=\frac{\pi}{2},\text{t}=1$
Therefore the integral becomes
$\text{I}=\int^\limits1_0\frac{2\text{dt}}{\text{a}-\text{a}\text{t}^2+2\text{bt}}$
$=\int^\limits1_0\frac{2\text{dt}}{-\text{a}\big[\text{t}^2-\frac{2\text{bt}}{\text{a}-1}\big]}$
$=\frac{2}{\text{a}}\int^\limits1_0\frac{\text{dt}}{-\Big[\big(\text{t}-\frac{\text{b}}{\text{a}}\big)^2-1-\frac{\text{b}^2}{\text{a}^2}\Big]}$
$=\frac{2}{\text{a}}\int^\limits1_0\frac{\text{dt}}{\Big(\frac{\text{b}^2}{\text{a}^2}+1\Big)-\big(\text{t}-\frac{\text{b}}{\text{a}}\big)^2}$
$=\frac{2}{\text{a}}\begin{bmatrix}\frac{1}{2\sqrt{\frac{\text{a}^2+\text{b}^2}{\text{a}^2}}}\begin{pmatrix}\log\begin{vmatrix}\frac{2\sqrt{\frac{\text{a}^2+\text{b}^2}{\text{a}^2}}+\big(\text{t}-\frac{\text{b}}{\text{a}}\big)}{\sqrt{\frac{\text{a}^2+\text{b}^2}{\text{a}^2}-\big(\text{t}-\frac{\text{b}}{\text{a}}\big)}}\end{vmatrix} \end{pmatrix}^1_0\end{bmatrix}$
$=\frac{1}{\sqrt{\text{a}^2+\text{b}^2}}\log\bigg(\frac{\text{a}+\text{b}+\sqrt{\text{a}^2+\text{b}^2}}{\text{a}+\text{b}-\sqrt{\text{a}^2+\text{b}^2}}\bigg)$

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