Question

please solve this...
and no fake answers ​

Answer 1

Question

$\sf \int \dfrac{1 + cos4x}{cosx - tanx} dx$

$\huge\bold{\gray{\sf{Answer:}}}$

$\bold{Explanation:}$

By using half angle formula:

$\sf= > cos4x = 2 {cos}^{2} 2x - 1$

$\sf= > 1 + cos4x = 2 {cos}^{2} x$

$\sf= > \displaystyle\int \dfrac{2 {cos}^{2}2x }{ \dfrac{cosx}{sinx} - \dfrac{sinx}{cosx} } dx$

$\sf= > \displaystyle \int \dfrac{2 {cos}^{2}2x }{ \dfrac{ {cos}^{2} x - {sin}^{2}x }{cosxsinx} } dx$

$\sf= > \displaystyle\int \dfrac{2 {cos}^{2}2xcosxsinx }{cos2x} dx$

$\sf= > \displaystyle\int \: cos2x(2sinxcosx)dx$

$\sf= > \dfrac{1}{2} \displaystyle\int \: 2cos2xsin2xdx$

$\sf= > \dfrac{1}{2} \displaystyle\int \: sin4xdx$

$\sf \boxed{\int \: Sinaxdx = - \dfrac{cosax}{a} + c}$

$\sf= > \dfrac{1}{2}\displaystyle \int \dfrac{(cos4x)}{4} + c = \dfrac{ - cos4x}{8} + c$

Answer 2

Hey Dear Brother.. Refer to Attachment for correct Answer..