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प्रश्न

\[\int\limits_0^{\pi/4} \sec x dx\]
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उत्तर

\[Let I = \int_0^\frac{\pi}{4} \sec x\ d\ x\ . Then, \]
\[I = \int_0^\frac{\pi}{4} \sec x \frac{\sec x + \tan x}{\sec x + \tan x} d x\]
\[ \Rightarrow I = \int_0^\frac{\pi}{4} \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x} d x\]
\[Put u = \sec x + \tan x\]
\[ \Rightarrow du = \sec^2 x + \sec x \tan x dx\]
\[ \therefore \int_0^\frac{\pi}{4} \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x} d x = \int\frac{du}{u}\]
\[ \Rightarrow I = \left[ \log u \right]\]
\[ \Rightarrow I = \left[ \log \left( \sec x + \tan x \right) \right]_0^\frac{\pi}{4} \]
\[ \Rightarrow I = \log \left( \sec\frac{\pi}{4} + \tan\frac{\pi}{4} \right) - \log \left( \sec 0 + \tan 0 \right)\]
\[ \Rightarrow I = \log (\sqrt{2} + 1) - \log 1\]
\[ \Rightarrow I = \log (\sqrt{2} + 1)\]

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अध्याय 19: Definite Integrals - Exercise 20.1 [पृष्ठ १६]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Exercise 20.1 | Q 12 | पृष्ठ १६

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