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Definite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals5 Marks
Question
Evaluate the following integrals:$\int_{0}^\limits{\pi}\frac{1}{3+2\sin\text{x}+\cos\text{x}}\text{ dx}$
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Answer

Let $\text{I}=\int_{0}^\limits{\pi}\frac{1}{3+2\sin\text{x}+\cos\text{x}}\text{ dx}$ Then,$\text{I}=\int_{0}^\limits{\pi}\frac{1}{3+2\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)+\frac{1-\tan^{2}\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\pi}\frac{1+\tan^{2}\frac{\text{x}}{2}}{2\tan^{2}\frac{\text{x}}{2}+4\tan\frac{\text{x}}{2}+4}\text{ dx}$
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}=\pi,\text{t}=\infty$$\therefore\ \text{I}=\int\limits^{\infty}_0\frac{2\text{dt}}{2\text{t}^2+4\text{t}+4}$
$\Rightarrow\text{I}=\int\limits^{\infty}_0\frac{\text{dt}}{(\text{t}+1)^2+1}$
$\Rightarrow\text{I}=\Big[\tan^{-1}\big(\text{t}+1\big)\Big]^{\infty}_0$
$\Rightarrow\text{I}=\frac{\pi}{2}-\frac{\pi}{4}$
$\Rightarrow\text{I}=\frac{\pi}{4}$

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