int {frac{{cos x}}{{cos (x - a)}}dx - } int { | vedclass

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(C) माना $I = \int \frac{\cos x}{\cos (x - a)} dx - \int \frac{\sin x}{\sin (x - a)} dx = I_1 - I_2$.$I_1 = \int \frac{\cos x}{\cos (x - a)} dx$ के लि

Vedclass · Accepted Answer

$\sin a \log |\cot (x - a)| + c$

Vedclass · Answer

(C) माना $I = \int \frac{\cos x}{\cos (x - a)} dx - \int \frac{\sin x}{\sin (x - a)} dx = I_1 - I_2$.$I_1 = \int \frac{\cos x}{\cos (x - a)} dx$ के लिए,$x = (x - a) + a$ प्रतिस्थापित करें:$I_1 = \int \frac{\cos((x - a) + a)}{\cos (x - a)} dx = \int \frac{\cos(x - a) \cos a - \sin(x - a) \sin a}{\cos (x - a)} dx$$I_1 = \cos a \int dx - \sin a \int 	an(x - a) dx = x \cos a + \sin a \log |\cos(x - a)| + C_1$.$I_2 = \int \frac{\sin x}{\sin (x - a)} dx$ के लिए,$x = (x - a) + a$ प्रतिस्थापित करें:$I_2 = \int \frac{\sin((x - a) + a)}{\sin (x - a)} dx = \int \frac{\sin(x - a) \cos a + \cos(x - a) \sin a}{\sin (x - a)} dx$$I_2 = \cos a \int dx + \sin a \int \cot(x - a) dx = x \cos a + \sin a \log |\sin(x - a)| + C_2$.अब,$I = I_1 - I_2 = (x \cos a + \sin a \log |\cos(x - a)|) - (x \cos a + \sin a \log |\sin(x - a)|) + C$$I = \sin a (\log |\cos(x - a)| - \log |\sin(x - a)|) + C$$I = \sin a \log \left| \frac{\cos(x - a)}{\sin(x - a)} ight| + C = \sin a \log |\cot(x - a)| + C$.

$\int {\frac{{\cos x}}{{\cos (x - a)}}dx - } \int {\frac{{\sin x}}{{\sin (x - a)}}dx = } $

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