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 {\frac{{dx}}{{7 + 5\cos x}} = } $
  • ✓
    $\frac{1}{{\sqrt 6 }}{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 6 }}\tan \frac{x}{2}} \right) + c$
  • B
    $\frac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}\tan \frac{x}{2}} \right) + c$
  • C
    $\frac{1}{4}{\tan ^{ - 1}}\left( {\tan \frac{x}{2}} \right) + c$
  • D
    $\frac{1}{7}{\tan ^{ - 1}}\left( {\tan \frac{x}{2}} \right) + c$
✓

Answer

Correct option: A.
$\frac{1}{{\sqrt 6 }}{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 6 }}\tan \frac{x}{2}} \right) + c$
a
(a)$I = \frac{{dx}}{{7 + 5\cos x}}$$ = \int {\frac{{dx}}{{7 + 5\,\left( {\frac{{1 - {{\tan }^2}(x/2)}}{{1 + {{\tan }^2}(x/2)}}} \right)}}} $

$ = \int {\frac{{{{\sec }^2}(x/2)\,dx}}{{7 + 7{{\tan }^2}(x/2) + 5 - 5{{\tan }^2}(x/2)}}} $

$ = \int {\frac{{{{\sec }^2}(x/2)\,dx}}{{12 + 2{{\tan }^2}(x/2)}}} $$ = \int {\frac{{\frac{1}{2}{{\sec }^2}(x/2)\,.dx}}{{6 + {{\tan }^2}(x/2)}}} $

$\tan \frac{x}{2} = t$ $⇒$  $\frac{1}{2}{\sec ^2}\frac{x}{2}dx = dt$

$I = \int {\frac{{dt}}{{{t^2} + ({{\sqrt {6)} }^2}}}} $ $ = \frac{1}{{\sqrt 6 }}{\tan ^{ - 1}}\frac{t}{{\sqrt 6 }} + c$

$ = \frac{1}{{\sqrt 6 }}{\tan ^{ - 1}}\left| {\frac{{\tan (x/2)}}{{\sqrt 6 }}} \right| + 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 $f\left( x \right) = \left\{ \begin{gathered}
  {\left( {x - 1} \right)^{\frac{1}{{2 - x}}}},\,\,\,x > 1,x \ne 2 \hfill \\
  k\,,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 2 \hfill \\ 
\end{gathered}  \right.$ The value of $k$ for which $f$ is continuous at $x\, = 2$ is
→
The adjoint of $\left[ {\begin{array}{*{20}{c}}1&1&1\\1&2&{ - 3}\\2&{ - 1}&3\end{array}} \right]$is
→
Let $A$ be a $3 \times 3$ matrix such that $A^T + 2A= I$ then $det\,(A^{-1}) $ is equal to
→
Choose the correct answer from the given four options.
Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5).
Let F = 4x + 6y be the objective function.
The Minimum value of F occurs at.
→
Solution of differential equation $2xy\frac{{dy}}{{dx}} = {x^2} + 3{y^2}$ is

(where $p$ is a constant)

→
The maximum value of $Z=4 x+y$ for a $\text{L.P.P.}$ whose feasible region is given below is:
Image
→
Let $a, b$ and $c$ denote the outcome of three independent rolls of a fair tetrahedral die, whose four faces are marked $1,2,3,4$. If the probability that $a x^2+b x+c=0$ has all real roots is $\frac{m}{n}$, $\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then $\mathrm{m}+\mathrm{n}$ is equal to ..........
→
The production of item $A$ is $x$ and the production of item $B$ is $y .$ If the corner points of the bounded feasible region are $(1,0),(2,0),(0,2)$ and $(0,1)$ then the maximum profit $z=2000 x+5000 y$ is $\ldots \ldots$
→
The linear inequalities or equations or restrictions on the variables of a linear programming problem are called:
→
${d \over {dx}}{\sin ^{ - 1}}(3x - 4{x^3}) = $
→