MCQ

MCQ 12 Marks

$\int \frac{\cos (2 x)-1}{\cos (2 x)+1} d x=$______.

A
tan x-x + c
B
x + tan x + c
C
x- tan x + c
D
- x - cot x + c

Answer

$x-\tan x+c$

$\begin{aligned} \int \frac{\cos 2 x-1}{\cos 2 x+1} d x & =\int \frac{-(1-\cos 2 x)}{(1+\cos 2 x)} d x \\ =-\int \frac{2 \sin ^2 x}{2 \cos ^2 x} d x & =-\int \tan ^2 x d x \\ & =-\int\left(\sec ^2 x-1\right) d x \\ & =-(\tan x-x)+c \\ & =x-\tan x+c\end{aligned}$

View full question & answer→

MCQ 22 Marks

$\int \frac{\cos (2 x)-1}{\cos (2 x)+1} d x=$ _____.

A
$\tan x-x+c$
B
$x+\tan x+c$
C
$x-\tan x+c$
D
$-x-\cot x+c$

Answer

x − tan x + C
$\int\left(\frac{\cos 2 x-1}{\cos 2 x+1}\right) d x$
$=\int\left(\frac{1-2 \sin ^2 x-1}{2 \cos ^2 x-1+1}\right) d x$
$=-\int \tan ^2 x d x$
$=-\int\left(\sec ^2 x-1\right) d x$
$=\int\left(1-\sec ^2 x\right) d x$
$=x-\tan x+C$

View full question & answer→

MCQ 32 Marks

$\int \frac{1}{1+\cos x} d x=$______.

A
$\tan \left(\frac{x}{2}\right)+c$
B
$2 \tan \left(\frac{x}{2}\right)+c$
C
$-\cot \left(\frac{x}{2}\right)+c$
D
$-2 \cot \left(\frac{x}{2}\right)+c$

View full question & answer→

MCQ 42 Marks

If $\int \frac{d x}{4 x^2-1}=A \log \left(\frac{2 x-1}{2 x+1}\right)+c,$ then $A= ........ $

A
$1$
B
$\frac{1}{2}$
C
$\frac{1}{3}$
✓
$\frac{1}{4}$

Answer

Correct option: D.

$\frac{1}{4}$

$\frac{1}{4}$
$\int \frac{d x}{4 x^2-1}=A \log \left(\frac{2 x-1}{2 x+1}\right)+c ...\ ($given$)$
$\therefore \int \frac{d x}{(2 x)^2-1}= A \log \left(\frac{2 x-1}{2 x+1}\right)+c$
$\therefore \frac{1}{2(1)(2)} \log \left(\frac{2 x-1}{2 x+1}\right)+c$
$= A \log \left(\frac{2 x-1}{2 x+1}\right)+c$
$\therefore A =\frac{1}{4}$

View full question & answer→

MCQ 52 Marks

$\int \frac{d x}{9 x^2+1}=\ldots \ldots .$

A
$\frac{1}{3} \tan ^{-1}(2 x)+ c$
B
$\frac{1}{3} \tan ^{-1} x+c$
C
$\frac{1}{3} \tan ^{-1}(3 x)+c$
D
$\frac{1}{3} \tan ^{-1}(6 x)+c$

Answer

coming soon

View full question & answer→

MCQ 62 Marks

$\int \frac{1}{x} \cdot \log x d x= $______.

A
$\log (\log x)+c$
B
$\frac{1}{2}(\log x)^2+c$
C
$2 \log x+c$
D
$\log x+c$

Answer

coming soon

View full question & answer→

Maths MCQ

Test yourself on this topic