_ ^ dx 4 ^2 x + 5 ^2 x = - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{4{{\sin }^2}x + 5{{\cos }^2}x}} = } $

A
$\frac{1}{{\sqrt 5 }}{\tan ^{ - 1}}\left( {\frac{{2\tan x}}{{\sqrt 5 }}} \right) + c$
B
$\frac{1}{{\sqrt 5 }}{\tan ^{ - 1}}\left( {\frac{{\tan x}}{{\sqrt 5 }}} \right) + c$
✓
$\frac{1}{{2\sqrt 5 }}{\tan ^{ - 1}}\left( {\frac{{2\tan x}}{{\sqrt 5 }}} \right) + c$
D
None of these

✓

Answer

Correct option: C.

$\frac{1}{{2\sqrt 5 }}{\tan ^{ - 1}}\left( {\frac{{2\tan x}}{{\sqrt 5 }}} \right) + c$

c
(c)$\int_{}^{} {\frac{{dx}}{{4{{\sin }^2}x + 5{{\cos }^2}x}}} = \int_{}^{} {\frac{{{{\sec }^2}x\,dx}}{{4{{\tan }^2}x + 5}} = \frac{1}{4}\int_{}^{} {\frac{{{{\sec }^2}x\,dx}}{{{{\tan }^2}x + \frac{5}{4}}}} } $
Put $\tan x = t \Rightarrow {\sec ^2}x\,dx = dt,$ then it reduces to
$\frac{1}{4}\int_{}^{} {\frac{{dt}}{{{t^2} + {{\left( {\frac{{\sqrt 5 }}{2}} \right)}^2}}} = \frac{2}{{4\sqrt 5 }}{{\tan }^{ - 1}}\left( {\frac{{2t}}{{\sqrt 5 }}} \right)} + c$
$ = \frac{1}{{2\sqrt 5 }}{\tan ^{ - 1}}\left( {\frac{{2\tan x}}{{\sqrt 5 }}} \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

If $\text{g(f(x))}=|\sin\text{x}|$ and $\text{f(g(x))}=(\sin\sqrt{\text{x}})^2,$ then

The corner points of the bounded feasible region determined by a system of linear constraints are $(0,3),(1,1)$ and $(3,0)$. Let $Z=p x+q y$, where $p, q>0$,. The condition on $p$ and $q$ so that the minimum of $Z$ occurs at $(3,0)$ and $(1,1)$ is

The area bounded by the curve $\text{y}=\sin\text{x}$ between the ordinates $\text{x}=0,\text{x}=\pi$ and the $x-$ axis is :

If A = [1 amp; 2], B = [3 amp; 4] then A + B =

$\cos[\tan^{-1}\{\sin(\cot^{-1}\text{x})\}]$ is equal to:

The solution curve of the differential equation $y \frac{d x}{d y}=x\left(\log _e x-\log _e y+1\right), x>0, y>0 \text { passing }$ through the point$(\mathrm{e}, 1)$ is

The function $y = a(1 - \cos x)$ is maximum when $x = $

If $x = a\left( {t - {1 \over t}} \right)\,,y = a$ $\left( {t + {1 \over t}} \right)$ then ${{dy} \over {dx}} = $

$f(x) = x + \sqrt {{x^2}} $ is a function from $R \to R$ , then $f(x)$ is

$\int_{}^{} {\frac{1}{{{x^2}\sqrt {1 + {x^2}} }}} \;dx = $