Definite Integration
IBP + Recursive
Grade 12

Question:

<p>Evaluate \(\displaystyle\int_0^{\pi/4}\sec^3 x\,dx\) [JEE Main 2019]</p>
\(\dfrac{\sqrt{2}+\ln(1+\sqrt{2})}{2}\)
\(\dfrac{\sqrt{2}}{2}\)
\(\ln(1+\sqrt{2})\)
\(\dfrac{\sqrt{2}-1}{2}\)

Step-by-Step Solution

Key Concept: Use reduction: \intsec^3x dx = (sec x tan x)/2 + (1/2)\intsec x dx = (sec x tan x)/2 + (1/2)ln|sec x + tan x|.
<div class='solution'> <p>IBP: \(\int\sec^3 x\,dx=\sec x\tan x-\int\sec x\tan^2 x\,dx=\sec x\tan x-\int\sec x(\sec^2 x-1)dx\)</p> <p>\(=\sec x\tan x-\int\sec^3 x\,dx+\int\sec x\,dx\)</p> <p>\(2\int\sec^3 x\,dx=\sec x\tan x+\ln|\sec x+\tan x|+C\)</p> <p>\(\int_0^{\pi/4}\sec^3 x\,dx=\frac{1}{2}[\sec x\tan x+\ln|\sec x+\tan x|]_0^{\pi/4}\)</p> <p>\(=\frac{1}{2}[(\sqrt{2}\cdot1+\ln(\sqrt{2}+1))-(1\cdot0+\ln 1)]=\frac{\sqrt{2}+\ln(1+\sqrt{2})}{2}\)</p>
Correct Answer: A

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