Question

If $\int \frac{\cos x}{\cos (x-\alpha )}dx=Ax+B\log \left|\cos \right(x-\alpha \left)\right|+c,$ then

• A. $A=\cos \alpha$
• B. $A=\sin \alpha$
• C. $B=\sin \alpha$
• D. $B=\cos \alpha$

Answer 1

Answer: (A), (C)

The correct options are
A $A=\cos \alpha$
C $B=\sin \alpha$

$\int \frac{\cos x}{\cos (x-\alpha )}dx$

Putting $x-\alpha =\theta$

$=\int \frac{\cos (\alpha +\theta )}{\cos \theta }d\theta$

$=\int \frac{\cos \alpha \cos \theta -\sin \alpha \sin \theta }{\cos \theta }d\theta$

$=\cos \alpha \int d\theta -\sin \alpha \int \frac{\sin \theta }{\cos \theta }d\theta$

$=\left(\cos \alpha \right)\theta +\sin \alpha \log |\cos \theta |+K$,where$K$is an arbitrary constant

$=\left(\cos \alpha \right)(x-\alpha )+\sin \alpha \log |\cos \theta |+K$

$=\left(\cos \alpha \right)x+\sin \alpha \log |\cos (x-\alpha )|+C$,where$C=K-\alpha \cos \alpha$is an arbltrary constant.

$\int \frac{\cos x}{\cos (c-\alpha )}dx=Ax+B\log \left|cos\right(x-\alpha \left)\right|+c,$

On comparing, we get

$\Rightarrow A=\cos \alpha ,B=\sin \alpha .$