Question

Let $I_1=\int \tan x\tan (ax+b)dx$ and $I_2=\int \cot x\cot (ax+b)dx$

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)

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