Let I1=∫tan⁡xtan⁡(ax+b)dx and I2=∫cot⁡xcot⁡(ax+b)dxColumn - IColumn - Ia. Value of I1 for a=1 is p.x−cot⁡bln⁡cos⁡(x−b)cos⁡x+Cb. Value I2 for a=1 is q.cot⁡bln⁡sin⁡xsin⁡(x+b)−x+Cc. Value of I1 for a=−1r.cot⁡bln⁡cos⁡xcos⁡(x+b)−x+Cd. Value of I2 for a=−1s.x+cot⁡bln⁡sin⁡xsin⁡(b−x)+C

a=1 I1=∫tan⁡(x)tan⁡(x+b)dx Now, tan⁡(x+b−x)=tan⁡(x+b)−tan⁡x1+tan⁡(x+b)tan⁡x i.e.1+tan⁡(x+b)tan⁡x=tan⁡(x+b)−tan⁡xtan⁡b so⁡I1=∫(tan⁡(x+b)−tan⁡x)tan⁡b−1dx = 1tan⁡b(ln⁡sec⁡(x+b)−ln⁡sec⁡x)−x+C =cot⁡b(ln⁡cosx-ln⁡cos⁡(x+b))−x+C =cot⁡bln⁡cos⁡xcos⁡(x+b)−x+C(B) a=1 I2=∫cot⁡xcot⁡(x+b)dx Now cot⁡(x+b−x)=cot⁡(x+b)cot⁡x+1cot⁡x−cot⁡(x+b) i.e. cot⁡(x+b)cot⁡x+1=cot⁡b(cot⁡x−cot⁡(x+b))I2=∫1tan⁡b(cot⁡x−cot⁡(x+b))−1dx=cot⁡b(ln⁡sin⁡x−ln⁡sin⁡(x+b))−x+C =cot⁡bln⁡sin⁡xsin⁡(x+b)−x+C(C) a = - 1 isi.e. I1=∫tan⁡xtan⁡(b−x)dxNow, tan⁡(x+b−x)=tan⁡x+tan⁡(b−x)1−tan⁡xtan⁡(b−x)i.e. 1−tan⁡xtan⁡(b−x)=tan⁡x+tan⁡(b−x)tan⁡bI1=∫1−1tan⁡b(tan⁡x+tan⁡(b−x))dx=x−1tan⁡b(lnsec⁡x−lnsec⁡(b−x))+C=x−1tan⁡b(lncos⁡(b−x)−lncos)+C=x−cot⁡bln⁡cos⁡(x−b)cos⁡x+C (D) a=−1 I1=∫cot⁡xcot⁡(b−x)dxNow, cot⁡(x+b−x)=cot⁡xcot⁡(b−x)−1cot⁡x+cot⁡(b−x)i.e. cot⁡xcot⁡(b−x)=1+cot⁡b(cot⁡x+cot⁡(b−x))I2=∫1+cot⁡b(cot⁡x+cot⁡(b−x))dx =x+cot⁡bln⁡sin⁡xsin⁡(b−x)+C

Let I1=∫tan⁡xtan⁡(ax+b)dx and I2=∫cot⁡xcot⁡(ax+b)dxColumn - IColumn - Ia. Value of I1 for a=1 is p.x−cot⁡bln⁡cos⁡(x−b)cos⁡x+Cb. Value I2 for a=1 is q.cot⁡bln⁡sin⁡xsin⁡(x+b)−x+Cc. Value of I1 for a=−1r.cot⁡bln⁡cos⁡xcos⁡(x+b)−x+Cd. Value of I2 for a=−1s.x+cot⁡bln⁡sin⁡xsin⁡(b−x)+C

a=1 I1=∫tan⁡(x)tan⁡(x+b)dx Now, tan⁡(x+b−x)=tan⁡(x+b)−tan⁡x1+tan⁡(x+b)tan⁡x i.e.1+tan⁡(x+b)tan⁡x=tan⁡(x+b)−tan⁡xtan⁡b so⁡I1=∫(tan⁡(x+b)−tan⁡x)tan⁡b−1dx = 1tan⁡b(ln⁡sec⁡(x+b)−ln⁡sec⁡x)−x+C =cot⁡b(ln⁡cosx-ln⁡cos⁡(x+b))−x+C =cot⁡bln⁡cos⁡xcos⁡(x+b)−x+C(B) a=1 I2=∫cot⁡xcot⁡(x+b)dx Now cot⁡(x+b−x)=cot⁡(x+b)cot⁡x+1cot⁡x−cot⁡(x+b) i.e. cot⁡(x+b)cot⁡x+1=cot⁡b(cot⁡x−cot⁡(x+b))I2=∫1tan⁡b(cot⁡x−cot⁡(x+b))−1dx=cot⁡b(ln⁡sin⁡x−ln⁡sin⁡(x+b))−x+C =cot⁡bln⁡sin⁡xsin⁡(x+b)−x+C(C) a = - 1 isi.e. I1=∫tan⁡xtan⁡(b−x)dxNow, tan⁡(x+b−x)=tan⁡x+tan⁡(b−x)1−tan⁡xtan⁡(b−x)i.e. 1−tan⁡xtan⁡(b−x)=tan⁡x+tan⁡(b−x)tan⁡bI1=∫1−1tan⁡b(tan⁡x+tan⁡(b−x))dx=x−1tan⁡b(lnsec⁡x−lnsec⁡(b−x))+C=x−1tan⁡b(lncos⁡(b−x)−lncos)+C=x−cot⁡bln⁡cos⁡(x−b)cos⁡x+C (D) a=−1 I1=∫cot⁡xcot⁡(b−x)dxNow, cot⁡(x+b−x)=cot⁡xcot⁡(b−x)−1cot⁡x+cot⁡(b−x)i.e. cot⁡xcot⁡(b−x)=1+cot⁡b(cot⁡x+cot⁡(b−x))I2=∫1+cot⁡b(cot⁡x+cot⁡(b−x))dx =x+cot⁡bln⁡sin⁡xsin⁡(b−x)+C

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$Let I_{1} = \int \tan x \tan (a x + b) d x and I_{2} = \int \cot x \cot (a x + b) d x$
Column - I Column - I
a. $Value of I_{1} for a = 1 is$ p. $x - \cot bln (\frac{\cos (x - b)}{\cos x}) + C$
b. $Value I_{2} for a = 1 is$ q. $\cot bln (\frac{\sin x}{\sin (x + b)}) - x + C$
c. $Value of I_{1} for a = - 1$ r. $\cot bln (\frac{\cos x}{\cos (x + b)}) - x + C$
d. $Value of I_{2} for a = - 1$ s. $x + \cot bln (\frac{\sin x}{\sin (b - x)}) + C$

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A-p, B-q, C-s, D-r

A-r, B-q, C-p, D-s

A-r, B-p, C-s, D-q

A-s, B-p, C-q, D-r

answer is B.

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Detailed Solution

$a = 1 I_{1} = \int \tan (x) \tan (x + b) dx Now, \tan (x + b - x) = \frac{\tan (x + b) - \tan x}{1 + \tan (x + b) \tan x} i . e . 1 + \tan (x + b) \tan x = \frac{\tan (x + b) - \tan x}{\tan b} so I_{1} = \int (\frac{(\tan (x + b) - \tan x)}{\tan b} - 1) d x = \frac{1}{\tan b} (\ln \sec (x + b) - \ln \sec x) - x + C = \cot b (\ln \cos x - \ln \cos (x + b)) - x + C = \cot bln (\frac{\cos x}{\cos (x + b)}) - x + C$

$\begin{array}{l} (B) a = 1 I_{2} = \int \cot xcot (x + b) dx \\ Now \cot (x + b - x) = \frac{\cot (x + b) \cot x + 1}{\cot x - \cot (x + b)} \\ i.e. \cot (x + b) \cot x + 1 = \cot b (\cot x - \cot (x + b)) \\ I_{2} = \int \frac{1}{\tan b} (\cot x - \cot (x + b)) - 1 dx \\ = \cot b (\ln \sin x - \ln \sin (x + b)) - x + C = \cot bln (\frac{\sin x}{\sin (x + b)}) - x + C \end{array}$

i.e. $I_{1} = \int \tan xtan (b - x) dx$

Now, $\tan (x + b - x) = \frac{\tan x + \tan (b - x)}{1 - \tan xtan (b - x)}$

i.e. $1 - \tan xtan (b - x) = \frac{\tan x + \tan (b - x)}{\tan b}$

$\begin{array}{l} I_{1} = \int 1 - \frac{1}{\tan b} (\tan x + \tan (b - x)) dx \\ = x - \frac{1}{\tan b} (lnsec x - lnsec (b - x)) + C \\ = x - \frac{1}{\tan b} (lncos (b - x) - lncos) + C \\ = x - \cot b (\ln \frac{\cos (x - b)}{\cos x}) + C \end{array}$

(D) $a = - 1 I_{1} = \int \cot xcot (b - x) dx$

Now, $\cot (x + b - x) = \frac{\cot xcot (b - x) - 1}{\cot x + \cot (b - x)}$

i.e. $\cot xcot (b - x) = 1 + \cot b (\cot x + \cot (b - x))$

$I_{2} = \int 1 + \cot b (\cot x + \cot (b - x)) dx = x + \cot bln (\frac{\sin x}{\sin (b - x)}) + C$

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