Download App

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

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.1 inderfinite integral1 Mark
MCQ
$\int_{}^{} {\sec x\log (\sec x + \tan x)\;dx = } $
  • A
    ${[\log (\sec x + \tan x)]^2} + c$
  • ✓
    $\frac{1}{2}{[\log (\sec x + \tan x)]^2} + c$
  • C
    ${\sec ^2}x + \tan x\sec x + c$
  • D
    None of these
✓

Answer

Correct option: B.
$\frac{1}{2}{[\log (\sec x + \tan x)]^2} + c$
b
(b) Let $\log (\sec x + \tan x) = t \Rightarrow \sec x\,dx = dt$
Therefore $\int_{}^{} {\sec x\,\log (\sec x + \tan x)\,dx} = \int_{}^{} {t\,dt} $
$ = \frac{{{t^2}}}{2} + c = \frac{{{{[\log (\sec x + \tan x)]}^2}}}{2} + 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

Explore more

More "M.C.Q (1 Marks)" from STD 12 - 7.1 inderfinite integral→STD 12 - 7.1 inderfinite integral — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

Let $O$ be the origin and $\overline{ OA }=2 \hat{ i }+2 \hat{ j }+\hat{ k }, \overline{ OB }=\hat{ i }-2 \hat{ j }+2 \hat{ k }$ and $\overline{ OC }=\frac{1}{2}(\overline{ OB }-\lambda \overline{ OA })$ for some $\lambda>0$. If $|\overline{ OB } \times \overline{ OC }|=\frac{9}{2}$, then which of the following statements is (are) TRUE?

$(A)$ Projection of $\overline{ OC }$ on $\overline{ OA }$ is $-\frac{3}{2}$

$(B)$ Area of the triangle $OAB$ is $\frac{9}{2}$

$(C)$ Area of the triangle $ABC$ is $\frac{9}{2}$

$(D)$ The acute angle between the diagonals of the parallelogram with adjacent sides $\overline{ OA }$ and $\overline{ OC }$ is $\frac{\pi}{3}$

→
The function $\text{f(x)}=\sin^{-1}(\cos\text{x})$ is:
→
If $\vec{\text{a}},\ \vec{\text{b}}$ are the vectors forming consecutive sides of a regular hexagon $\text{ABCDEF},$ then the vector representing side $CD$ is,
→
The number of real solution of the equation
$\sqrt{1+\cos2\text{x}}=\sqrt2\sin^{-1}(\sin\text{x}),-\pi\leq\text{x}\leq\pi$ is:
→
${\tan ^{ - 1}}\left( {\frac{1}{4}} \right) + {\tan ^{ - 1}}\left( {\frac{2}{9}} \right) = $
→
Let $\text{f}:\text{R}-\Big\{-\frac{4}{3}\Big\}\rightarrow\text{R}$ be a function defined as $f(\text{x})=\frac{4\text{x}}{3\text{x}+4}.$ The inverse of f is the map g: Range $\text{f}:\text{R}-\Big\{-\frac{4}{3}\Big\}\rightarrow\text{R}$ given by:
→
Let $f(x) = 2x^3 - 3x^2 - 12x + 5$ on $[-2, 4].$ The relative maximum occurs at $x =$
→
If $f(x) = \sin x - {x \over 2}$ is increasing function, then
→
Let  $f(x)\,\, = \,\left\{ {\begin{array}{*{20}{c}}
{x\,\sin \,\frac{1}{x}\,\sin \,\left( {\frac{1}{{x\,\sin \,\frac{1}{x}}}} \right)\,,\,\,x\, \ne \,0}\\
{0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\, = \,0\,}
\end{array}} \right.{\mkern 1mu} $  then $f(x)$ is
→
The value of $ \cos^{-1} (\cos 12) - \sin^{-1} (\sin 14)$ is
→