Question

$\int_{}^{}\frac{\cos 2x - \cos 2\theta}{\cos x - \cos\theta}$dx =

• A. 2(sin x + x cos q) + c
• B. 2(sin x – x cos q) + c
• C. 2 (sin x + 2x cos q) + c
• D. None of these

Answer 1

Answer: (A)

2(sin x + x cos q) + c

Answer 2

$\int \frac { \cos 2 x - \cos 2 \theta } { \cos x - \cos \theta } d x$

= $\int \frac { \left( 2 \cos ^ { 2 } x - 1 \right) - \left( 2 \cos ^ { 2 } \theta - 1 \right) } { \cos x - \cos \theta } d x$ = $2 \int \frac { \cos ^ { 2 } x - \cos ^ { 2 } \theta } { \cos x - \cos \theta } d x$

= $2 \int ( \cos x - \cos \theta ) d x$

= 2(sin x + x cos q) + c