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Definite Integrals — Maths STD 12 Science — Question

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

Let $\int_{0}^\limits{\frac{\pi}{4}}\sec\text{x}\text{ dx}$ Then,$\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\sec\text{x}\frac{\sec\text{x}+\tan\text{x}}{\sec\text{x}+\tan\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\frac{\sec^2\text{x}+\sec\text{x}\tan\text{x}}{\sec\text{x}+\tan\text{x}}\text{ dx}$
Put $\text{u}=\sec\text{x}+\tan\text{x}$$\Rightarrow\text{du}=\sec^2\text{x}+\sec\text{x}\tan\text{x dx}$
$\therefore\ \int_{0}^\limits{\frac{\pi}{4}}\frac{\sec^2\text{x}+\sec\text{x}\tan\text{x}}{\sec\text{x}+\tan\text{x}}\text{ dx}=\int\frac{\text{du}}{\text{u}}$
$\Rightarrow\text{I}=\big[\log\text{u}\big]$
$\Rightarrow\text{I}=\big[\log(\sec\text{x}+\tan\text{x})\big]^{\frac{\pi}{4}}_0$
$\Rightarrow\text{I}=\log\Big(\sec\frac{\pi}{4}+\tan\frac{\pi}{4}\Big)-\log(\sec0+\tan0)$
$\Rightarrow\text{I}=\log\big(\sqrt{2}+1\big)-\log1$
$\Rightarrow\text{I}=\log\big(\sqrt{2}+1\big)$

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