MCQ
$\int \frac{\sec x}{\sec x-\tan x} d x \text { equals }$
  • A
    $\sec x-\tan x+c$
  • B
    $\sec x+\tan x+c$
  • $\tan x-\sec x+c$
  • D
    $-(\sec x+\tan x)+c$

Answer

Correct option: C.
$\tan x-\sec x+c$
$\text {Let } I=\int \frac{\sec x}{\sec x-\tan x} d x$
$=\int \frac{\sec x(\sec x+\tan x)}{(\sec x-\tan x)(\sec x+\tan x)} d x=\int\left(\frac{\sec ^2 x+\sec x \tan x}{\sec ^2 x-\tan ^2 x}\right) d x$
$=\int \sec ^2 x d x+\int \sec x \tan x d x \quad\left[\because \sec ^2 x-\tan ^2 x=1\right]$
$=\tan x+\sec x+c$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Range of $\text{f(x)}=\sqrt{(1-\cos\text{x})\sqrt{(1-\cos\text{x})\sqrt{(1-\cos\text{x}).....\infty}}}$
Which of the following transformation reduce the differential quation into the form $\frac{\text{du}}{\text{dx}}+\text{P}(\text{x})\text{u}=\text{Q}(\text{x})$ into the from $\frac{\text{dz}}{\text{dx}}+\frac{\text{z}}{\text{x}}\log\text{z}=\frac{\text{z}}{\text{x}^{2}}(\log\text{z})^{2}$
The motion of stone thrown up vertically is given by $s = 13.8t - 4.9{t^2}$, where $s$ is in metre and  $t $ is in seconds. Then its velocity at $t = 1$ second is ........ $m/s$
Let matrix $A = \left[ {\begin{array}{*{20}{c}}
  1&2&3 \\ 
  0&5&4 \\ 
  0&3&2 
\end{array}} \right]$ and $A^3 -8A^2 + \alpha A + \beta I = O$ then ordered pair $(\alpha , \beta)$ is
Consider the system of equations: $a_1x + b_1y + c_1z = 0 , a_2x + b_2y + c_2z = 0 , a_3x + b_3y + c_3z = 0,$
if$\begin{vmatrix}\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2\\\text{a}_3&\text{b}_3&\text{c}_3\end{vmatrix}=0$, then the system has
The solution of the differential equation ${\sec ^2}x\tan ydx + {\sec ^2}y\tan xdy = 0$ is
$D$ is a $3 \times 3$ diagonal matrix. Which of the following statements is not true?
Function $f(x)={\left( {1 + \frac{1}{x}} \right)^x}$ then Domain of $f (x)$ is
Let $\vec a$ and $\vec b$ be two unit vectors such that $\left| {\vec a\, + \,\vec b} \right| = \sqrt 3 .$ If $\vec c = \vec a\, + \,2\vec b + 3\,(\vec a \times \vec b),$ then $2\left| {\vec c} \right|$ is equal to
The area of the region (in the square unit) bounded by the curve ${x^2} = 4y,$ line $x = 2$ and $x-$ axis is