Question

Let $I_1=\int \tan x \tan (a x+b) d x$ and $I_2=\int \cot x \cot (a x+b) d x$

Column I		Column II
A	value of $I _1$ for $a =1$ is	P	$ x-\cot b \ln \frac{\cos (x-b)}{\cos x}+C$
B	value of $I _2$ for $a =1$ is	Q	$ \cot b \ln \frac{\sin x}{\sin (x+b)}-x+C$
C	value of $I _1$ for $a =-1$ is	R	$ \cot b \ln \left(\frac{\cos x}{\cos (x+b)}\right)-x+C$
D	value of $I _2$ for $a =-1$ is	S	$ x+\cot b \ln \left(\frac{\sin x}{\sin (b-x)}\right)+C$

• A. (A) R ; (B) Q ; (C) P ; (D) P
• B. (A) R ; (B) Q ; (C) P ; (D) Q
• C. (A) R ; (B) Q ; (C) P ; (D) R
• D. (A) R ; (B) Q ; (C) P ; (D) S

Answer 1

Answer: (D)

$a=1 I_1=\int \tan (x) \tan (x+b) d x$
Now $\tan (x+b-x)=\frac{\tan (x+b)-\tan x}{1+\tan (x+b) \tan x}$
i.c. $1+\tan (x+b) \tan x=\frac{\tan (x+b)-\tan x}{\tan b}$
so $ I_1=\int \frac{1}{\tan b}(\tan (x+b)-\tan x)-1 d x$
i.e. $\frac{1}{\tan b}(\ln \sec (x+b)-\ln \sec x)-x+C$
i.c. $ \cot b (\ln \cos x -\ln \cos ( x + b ))- x + C$
(B) $ a =1 I _2=\int \cot x \cot ( x + b ) dx$
$\text { Now } \cot (x+b-x)=\frac{\cot (x+b) \cot x+1}{\cot x-\cot (x+b)} $
$\text { i.e. } \cot (x+b) \cot x+1=\cot b(\cot x-\cot (x+b)) $
$I_2=\int \frac{1}{\tan b}(\cot x-\cot (x+b))-1 d x $
$=\cot b(\ln \sin x-\ln \sin (x+b))-x+C$
(C) $I _1=$ for $a =-1$ is
i.c. $I_1=\int \tan x \tan (b-x) d x$
Now $\tan (x+b-x)=\frac{\tan x+\tan (b-x)}{1-\tan x \tan (b-x)}$
i.c. $ 1-\tan x \tan (b-x)=\frac{\tan x+\tan (b-x)}{\tan b}$
$I_1=\int 1-\frac{1}{\tan b}(\tan x+\tan (b-x)) d x $
$=x-\frac{1}{\tan b}(\ln \sec x-\ln \sec (b-x))+C$
$=x-\frac{1}{\tan b}(\ln \cos (b-x)-\ln \cos x)+C$
$=x-\cot b\left(\ln \frac{\cos (x-b)}{\cos x}\right)+C$
(D) $ I _2$ for $a =-1$
$I_1=\int \cot x \cot (b-x) d x$
Now $\cot (x+b-x)=\frac{\cot x \cot (b-x)-1}{\cot x+\cot (b-x)}$
i.e. $\cot x \cot (b-x)=1+\cot b(\cot x+\cot (b-x))$
$I _2=\int 1+\cot b (\cot x +\cot ( b - x )) dx $
$= x +\cot b (\ln \sin x -\ln \sin ( b - x ))+ C $