Question

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

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

Answer 1

Answer: (A), (B)

$A=\cos \alpha$
$B=\sin \alpha$

Given,

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

Let $x-\alpha =z$

or, $x=z+\alpha$ and $dx=dz$

Using these in above we get,

$=\int \frac{\cos (z+\alpha )}{\cos z}dz$

$=\int \frac{\cos z.\cos \alpha -\sin z.\sin \alpha }{\cos z}dz$

$=\int \cos \alpha dz-\int \tan z\sin \alpha dz$

$=z\cos \alpha +\sin \alpha .\log |\cos z|+c$ [Where $c$ is integrating constant]

$=(x-\alpha )\cos \alpha +\sin \alpha .\log |\cos (x-\alpha )|+c$

Now comparing this with the given expression we get,

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