Question

Integrate 1/2-3 cos 2x.

Answer 1

Hi Please find the attachment
Hope this helps.

Answer 2

Answer:

$\int \: \frac{1}{2 - 3cos \: 2x} dx =\frac{1}{2 \sqrt{5} } log\bigg( \bigg| \frac{ \sqrt{5} \: tan \: x - 1 }{ \sqrt{5} \: tan \: x + 1 }\bigg | \bigg) + c$

Step-by-step explanation:

To integrate

$\int \: \frac{1}{2 - 3cos \: 2x} dx \\ \\ \int \: \frac{1}{2 - 3(2 {cos}^{2} x - 1)} dx \\ \\ \int \: \frac{1}{2 - 6 {cos}^{2} x + 3} dx \\ \\ \int \: \frac{1}{5 - 6 {cos}^{2} x } dx$

or

$\int \: \frac{ {sec}^{2}x }{5 {sec}^{2} x - 6} dx \\ \\ \int \: \frac{ {sec}^{2}x }{5(1 + { tan}^{2} x) - 6} dx \\ \\ \int \: \frac{ {sec}^{2}x }{5 { tan}^{2} x - 1} dx$

let

$tan \: x = t \\ \\ {sec}^{2}x dx = dt$

Substitute the value

$\int \frac{dt}{5 {t}^{2} - 1 } \\ \\ = \frac{1}{5} \int \frac{dt}{ {t}^{2} - \big( { \frac{1}{ \sqrt{5} } \big)}^{2} }$

We know that

$\boxed{ \int \frac{1}{ {x}^{2} - {a}^{2} } dx = \frac{1}{2a} log\bigg( \bigg| \frac{x - a}{x + a} \bigg| \bigg) + c}$

$\frac{1}{5} \int \frac{dt}{ {t}^{2} - ( { \frac{1}{ \sqrt{5} } )}^{2} } \\ \\ = \frac{1}{5} \frac{1}{2 \frac{1}{ \sqrt{5} } } log\bigg( \bigg| \frac{t - \frac{1}{ \sqrt{5} } }{t + \frac{1}{ \sqrt{5} } } \bigg | \bigg) + c \\ \\ = \frac{ \sqrt{5} }{10} log\bigg( \bigg| \frac{ \sqrt{5}t - 1 }{ \sqrt{5}t + 1 } \bigg | \bigg) + c \\ \\ = \frac{1}{2 \sqrt{5} } log\bigg( \bigg| \frac{ \sqrt{5} \: tan \: x - 1 }{ \sqrt{5} \: tan \: x + 1 }\bigg | \bigg) + c$

Hope it helps you