Question

integrate dx/3+cosx​

Answer 1

Answer:

we know that

$cos \: 2x = \frac{ 1 - {tan}^{2}x }{1 + {tan}^{2} x}$

so integration of

$\frac{dx}{3 + \frac{1 - {tan}^{2} \frac{x}{2} }{1 + {tan}^{2} \frac{x}{2} } } = \frac{ {sec}^{2} \frac{x}{2}dx }{4 + 2 {tan}^{2} \frac{x}{2} } .........(i)$

Let

$\tan( \frac{x}{2} ) = t$

$\frac{1}{2} {sec}^{2} ( \frac{x}{2} )dx = dt$

from (i), integration of

$\frac{2dt}{4 + 2 {t}^{2} } = \frac{dt}{2 + {t}^{2} } =$

$\frac{1}{ \sqrt{2} } \tan {}^{ - 1} ( \frac{t}{ \sqrt{2} } ) + c$

replacing t, we get

$\frac{1}{ \sqrt{2} } {tan}^{ - 1} (\frac{ \tan( \frac{x}{2} ) }{ \sqrt{2} } ) + c$

Ans.