First slide

Introduction to integration

Question

$If f^{'} (x) = \tan^{- 1} (\sec x + \tan x), - \frac{π}{2} < x < \frac{π}{2}, and f (0) = 0, then f (1) is equal to:$

Difficult

A

$\frac{π + 2}{4}$
B

$\frac{π - 1}{4}$
C

$\frac{1}{4}$
D

$\frac{π + 1}{4}$

Solution

$f^{'} (x) = \tan^{- 1} (\sec x + \tan x)$
$\begin{array}{l} = \tan^{- 1} (\frac{1 + \sin x}{\cos x}) \\ = \tan^{- 1} (\frac{1 - \cos (\frac{π}{2} + x)}{\sin (\frac{π}{2} + x)}) \\ = \tan^{- 1} (\frac{2 \sin^{2} (\frac{π}{4} + \frac{x}{2})}{2 \sin (\frac{π}{4} + \frac{x}{2}) \cos (\frac{π}{4} + \frac{x}{2})}) \end{array}$
$= \tan^{- 1} (\tan (\frac{π}{4} + \frac{x}{2})) = \frac{π}{4} + \frac{x}{2} \int f^{'} (x) d x = \int (\frac{π}{4} + \frac{x}{2}) d x f (x) = \frac{π}{4} x + \frac{x^{2}}{4} + c$
$\begin{array}{l} f (0) = c = 0 \Rightarrow f (x) = \frac{π}{4} x + \frac{x^{2}}{4} \\ So f (1) = \frac{π + 1}{4} \end{array}$

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