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STD 12 - 7.1 inderfinite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.1 inderfinite integral1 Mark
MCQ
$\int_{}^{} {\frac{{\tan x}}{{\sec x + \tan x}}\;dx = } $
  • A
    $\sec x + \tan x - x + c$
  • ✓
    $\sec x - \tan x + x + c$
  • C
    $\sec x + \tan x + x + c$
  • D
    $ - \sec x - \tan x + x + c$
✓

Answer

Correct option: B.
$\sec x - \tan x + x + c$
b
(b)$\int_{}^{} {\frac{{\tan x}}{{(\sec x + \tan x)}}\,dx} = \int_{}^{} {\frac{{\tan x(\sec x - \tan x)}}{{(\sec x + \tan x)(\sec x - \tan x)}}\,dx} $
Multiplying ${N^r}$ and $D'$ by $(\sec x - \tan x),$ we get
$ = \int_{}^{} {\frac{{\tan x(\sec x - \tan x)}}{{({{\sec }^2}x - {{\tan }^2}x)}}\,dx} = \int_{}^{} {(\sec x\tan x - {{\tan }^2}x)\,dx} $
$ = \int_{}^{} {\sec x\tan x\,dx} - \int_{}^{} {({{\sec }^2}x - 1)\,dx} $
$ = \int_{}^{} {\sec x\tan x\,dx} - \int_{}^{} {{{\sec }^2}x\,dx} + \int_{}^{} {1\,dx} $
$ = \sec x - \tan x + x + c$.
Trick : By inspection,
$\frac{d}{{dx}}\left\{ {\sec x + \tan x} \right\} = \sec x\tan x + {\sec ^2}x$
$ = \sec x(\sec x + \tan x) = \frac{{\sec x}}{{\sec x - \tan x}}$
$ \Rightarrow \frac{d}{{dx}}\left\{ {\sec x - \tan x + x + c} \right\} = \sec x\tan x - {\sec ^2}x + 1$
$ = - \sec x(\sec x - \tan x) + 1 = \frac{{ - \sec x}}{{\sec x + \tan x}} + 1 = \frac{{\tan x}}{{\sec x + \tan x}}$.

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