Question

integrate as soon as possible
cos2x/sin7x*cos5x​

Answer 1

Given to evaluate,

$\displaystyle\longrightarrow I=\int\dfrac{\cos(2x)}{\sin(7x)\cos(5x)}\ dx$

$\displaystyle\longrightarrow I=\int\dfrac{\cos(7x-5x)}{\sin(7x)\cos(5x)}\ dx$

Since $\cos(a-b)=\cos a\cos b-\sin a\sin b,$

$\displaystyle\longrightarrow I=\int\dfrac{\cos(7x)\cos(5x)-\sin(7x)\sin(5x)}{\sin(7x)\cos(5x)}\ dx$

$\displaystyle\longrightarrow I=\int\left(\dfrac{\cos(7x)\cos(5x)}{\sin(7x)\cos(5x)}-\dfrac{\sin(7x)\sin(5x)}{\sin(7x)\cos(5x)}\right)\ dx$

$\displaystyle\longrightarrow I=\int\left(\cot(7x)-\tan(5x)\right)\ dx$

$\displaystyle\longrightarrow I=\int\cot(7x)\ dx-\int\tan(5x)\ dx\quad\quad\dots(1)$

Evaluating each integral,

$\displaystyle\longrightarrow\int\cot(7x)\ dx=-\int\dfrac{-\csc(7x)\cot(7x)}{\csc(7x)}\ dx$

$\displaystyle\longrightarrow\int\cot(7x)\ dx=-\dfrac{1}{7}\int\dfrac{-\csc(7x)\cot(7x)}{\csc(7x)}\ d(7x)$

$\displaystyle\longrightarrow\int\cot(7x)\ dx=-\dfrac{1}{7}\ln|\csc(7x)|$

And,

$\displaystyle\longrightarrow\int\tan(5x)\ dx=\int\dfrac{\sec(5x)\tan(5x)}{\sec(5x)}\ dx$

$\displaystyle\longrightarrow\int\tan(5x)\ dx=\dfrac{1}{5}\int\dfrac{\sec(5x)\tan(5x)}{\sec(5x)}\ d(5x)$

$\displaystyle\longrightarrow\int\tan(5x)\ dx=\dfrac{1}{5}\ln|\sec(5x)|$

Then (1) becomes,

$\displaystyle\longrightarrow\underline{\underline{I=-\dfrac{1}{7}\ln|\csc(7x)|-\dfrac{1}{5}\ln|\sec(5x)|+C}}$