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Question: \(\int\) [1 + tan x . tan (x + a)] dx is equal to –...

∫\int [1 + tan x . tan (x + a)] dx is equal to –

A

cot a . log∣sin⁡xsin⁡(x+α)∣+C\left| \frac{\sin x}{\sin(x + \alpha)} \right| + C

B

tan a . log∣sin⁡xsin⁡(x+α)∣+C\left| \frac{\sin x}{\sin(x + \alpha)} \right| + C

C

cota . log∣cos⁡xcos⁡(x+α)∣+C\left| \frac{\cos x}{\cos(x + \alpha)} \right| + C

D

none of these

Answer

cota . log∣cos⁡xcos⁡(x+α)∣+C\left| \frac{\cos x}{\cos(x + \alpha)} \right| + C

Explanation

Solution

∫{1+sin⁡xsin⁡(x+α)cos⁡x⋅cos⁡(x+α)}\int \left\{ 1 + \frac { \sin x \sin ( x + \alpha ) } { \cos x \cdot \cos ( x + \alpha ) } \right\} dx

=

= = cos⁡α∫dxcos⁡x⋅cos⁡(x+α)\cos \alpha \int \frac { d x } { \cos x \cdot \cos ( x + \alpha ) }

Multiply denominator and numerator by

sin a= cos⁡αsin⁡α∫sin⁡αcos⁡x⋅cos⁡(x+α)dx\frac { \cos \alpha } { \sin \alpha } \int \frac { \sin \alpha } { \cos x \cdot \cos ( x + \alpha ) } d x

= cot⁡α∫sin⁡(x+α−x)cos⁡x⋅cos⁡(x+α)dx\cot \alpha \int \frac { \sin ( x + \alpha - x ) } { \cos x \cdot \cos ( x + \alpha ) } d x

= cot⁡α∫sin⁡(x+α)cos⁡x−cos⁡(x+α)sin⁡xcos⁡x⋅cos⁡(x+α)dx\cot \alpha \int \frac { \sin ( x + \alpha ) \cos x - \cos ( x + \alpha ) \sin x } { \cos x \cdot \cos ( x + \alpha ) } d x

=

=

= =