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Question Number 43551 by peter frank last updated on 11/Sep/18

$$evaluate\int \frac{1}{\cos (x-a)\cos (x-b)}dx$$

Answered by MJS last updated on 12/Sep/18

$$\int \frac{dx}{\cos (x-a)\cos (x-b)}=$$$$[t=x-a\to dx=dt]$$$$=\int \frac{dt}{\cos t\cos (t+a-b)}=$$$$[\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta ]$$$$=\int \frac{dt}{\cos t(\cos t\cos (a-b)-\sin t\sin (a-b))}=$$$$[\cos (a-b)=p;\sin (a-b)=q]$$$$=\int \frac{dt}{p{\cos }^{2}t+q\cos t\sin t}=$$$$=\int \frac{dt}{{\cos }^{2}t(p+q\tan t)}=\int \frac{{\sec }^{2}t}{p+q\tan t}dt=$$$$[u=\tan t\to dt=\frac{du}{{\sec }^{2}t}]$$$$=\int \frac{du}{p+qu}=$$$$[v=p+qu\to du=\frac{dv}{q}]$$$$=\frac{1}{q}\int \frac{dv}{v}=\frac{1}{q}\ln v=\frac{1}{q}\ln (p+qu)=\frac{1}{q}\ln (p+q\tan t)=$$$$=\frac{1}{\sin (a-b)}\ln \mid \cos (a-b)+\sin (a-b)\tan (x-a)\mid +C$$

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