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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $I = \int \frac{\cos x}{\cos (x - a)} dx - \int \frac{\sin x}{\sin (x - a)} dx = I_1 - I_2$.For $I_1 = \int \frac{\cos x}{\cos (x - a)} dx$,su

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int \frac{\cos x}{\cos (x - a)} dx - \int \frac{\sin x}{\sin (x - a)} dx = I_1 - I_2$.For $I_1 = \int \frac{\cos x}{\cos (x - a)} dx$,substitute $x = (x - a) + a$:$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$$I_1 = \cos a \int dx - \sin a \int 	an(x - a) dx = x \cos a + \sin a \log |\cos(x - a)| + C_1$.For $I_2 = \int \frac{\sin x}{\sin (x - a)} dx$,substitute $x = (x - a) + a$:$I_2 = \int \frac{\sin((x - a) + a)}{\sin (x - a)} dx = \int \frac{\sin(x - a) \cos a + \cos(x - a) \sin a}{\sin (x - a)} dx$$I_2 = \cos a \int dx + \sin a \int \cot(x - a) dx = x \cos a + \sin a \log |\sin(x - a)| + C_2$.Now,$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 = } $

Similar Questions

$\int \left(\sum_{r=0}^{\infty} \frac{x^r 3^r}{r!}\right) dx =$

$\int {\frac{{{x^2}dx}}{{{{(a + bx)}^2}}}} = $

Let $f(x)=\int\left(\frac{2 x^3-3 x^2+4 x-5}{x^2}\right) d x$ and $f(1)=1$. Then $f(5)=$

$\int \frac{1}{x\sqrt{x^2 - 1}} \, dx = $

$\int \frac{1 - x^7}{x(1 + x^7)} dx$ equals:

Vedclass Test Series

Exam Paper Generator

Online Exam Module