MCQ
$\int\text{x}\sec\text{x}^2\text{ dx}$ is equal to:
  • $\frac{1}{2}\log\big(\sec\text{x}^2+\tan\text{x}^2\big)+\text{C}$
  • B
    $\frac{\text{x}^2}{2}\log\big(\sec\text{x}^2+\tan\text{x}^2\big)+\text{C}$
  • C
    $2\log\big(\sec\text{x}^2+\tan\text{x}^2\big)+\text{C}$
  • D
    none of these.

Answer

Correct option: A.
$\frac{1}{2}\log\big(\sec\text{x}^2+\tan\text{x}^2\big)+\text{C}$
$\text{I}=\int\text{x}\sec\text{x}^2\text{ dx}$
Put $\text{x}^2=\text{t}$
$=\text{x}=\sqrt{\text{t}}$
$2\text{xdx}=\text{dt}$
$\text{xdx}=\frac{\text{dt}}{2}$
$\text{I}=\int\sec\text{t}\frac{\text{dt}}{2}$
$\text{I}=\frac{1}{2}\log(\sec\text{t}+\tan\text{t})+\text{C}$
$\frac{1}{2}\log\big(\sec\text{x}^2+\tan\text{x}^2\big)+\text{C}$

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

Let $f\left( x \right) = \left\{ \begin{gathered}
  {\left( {x - 1} \right)^{\frac{1}{{2 - x}}}},\,\,\,x > 1,x \ne 2 \hfill \\
  k\,,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 2 \hfill \\ 
\end{gathered}  \right.$ The value of $k$ for which $f$ is continuous at $x\, = 2$ is
For $n \in N$ , let ${P_n} = \int\limits_1^e {{{\left( {\ln x} \right)}^n}dx} $ , then $(P_{10} -90P_8)$ is equal to
In Graphical solution the feasible solution is any solution to a $\text{LPP}$ which satisfies.
$\int\limits_0^{^n{C_r}} {\{ {{\sin }^2}\{ x\} \} dx} $ is equal to $($ where $\{.\}$ denotes fractional part function & $n, r  \in  N$ $)$
Let $\overrightarrow{P R}=3 \hat{i}+\hat{j}-2 \hat{k}$ and $\overrightarrow{S Q}=\hat{i}-3 \hat{j}-4 \hat{k}$ determine diagonals of a parallelogram $P Q R S$ and $\overrightarrow{P T}=\hat{i}+2 \hat{j}+3 \hat{k}$ be another vector. Then the volume of the parallelepiped determined by the vectors $\overrightarrow{ PT }, \overrightarrow{ PQ }$ and $\overrightarrow{ PS }$ is
If the position vectors of the point $A, B, C$  be $i, j, k $ respectively and $P $ be a point such that $\overrightarrow {AB} = \overrightarrow {CP} ,$ then the position vector of $P $ is
A wire of length $22 \;m$ is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then, the length of the side of the equilateral triangle, so that the combined area of the square and the equilateral triangle is minimum, is
Let $A = \{1, 2, 3\}$ and consider the relation $R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\}.$ Then $R$ is:
Let $[x]$ denote the integral part of $x \in R.$ $g(x) = x - [x]$. Let $f(x)$ be any continuous function with $f(0) = f(1)$ then the function $h(x) = f(g(x))$ :
$\int_{}^{} {\frac{{x{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}\;} dx = $