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प्रश्न
\[\int\frac{\cos x}{\cos \left( x - a \right)} dx\]
योग
उत्तर
\[\text{Let I} = \int\frac{\ cosx}{\cos\left( x - a \right)}dx\]
\[\text{Putting x }- a = t \]
\[ \Rightarrow x = a + t\]
\[ \Rightarrow dx = dt\]
\[ \therefore I = \int\frac{\cos\left( a + t \right)dt}{\cos t }\]
\[ = \int\frac{\cos a \cos t}{\cos t} - \frac{\sin a \sin t}{\cos t}dt\]
\[ = \int\left( \cos a - \sin a \tan t \right)dt\]
\[ = t\cos a - \text{sin a } In \left| \text{sec t} \right| + C\]
\[ = \left( x - a \right)\cos a - \text{sin a } In\left| \sec\left( x - a \right) \right| + C \left[ \because t = x - a \right]\]
\[\text{Putting x }- a = t \]
\[ \Rightarrow x = a + t\]
\[ \Rightarrow dx = dt\]
\[ \therefore I = \int\frac{\cos\left( a + t \right)dt}{\cos t }\]
\[ = \int\frac{\cos a \cos t}{\cos t} - \frac{\sin a \sin t}{\cos t}dt\]
\[ = \int\left( \cos a - \sin a \tan t \right)dt\]
\[ = t\cos a - \text{sin a } In \left| \text{sec t} \right| + C\]
\[ = \left( x - a \right)\cos a - \text{sin a } In\left| \sec\left( x - a \right) \right| + C \left[ \because t = x - a \right]\]
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