Question

integration of 1/(3sin2x+4cos2x)
note: 1/(3 sin2x+ 4 cos 3x) it is .there is no square​​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\displaystyle\int\sf \: \dfrac{1}{3sin2x \: + \: 4cos2x} \: dx$

We know, that

$\green{ \boxed{ \bf \: sin2x = \frac{2tanx}{1 + {tan}^{2}x }}}$

and

$\green{ \boxed{ \bf \: cos2x = \frac{1 - {tan}^{2}x }{1 + {tan}^{2}x }}}$

So, on substituting these Identities, we get

$\rm  \:  =  \: \displaystyle\int\sf\:\dfrac{1}{3 \times \dfrac{2tanx}{1 + {tan}^{2} x} + 4 \times \dfrac{1 - {tan}^{2} x}{1 + {tan}^{2} x}} \: dx$

$\rm  \:  =  \: \:\displaystyle\int\sf \: \dfrac{1 + {tan}^{2}x }{6tanx + 4 - 4 {tan}^{2} x} \: dx$

$\rm  \:  =  \: - \:\displaystyle\int\sf \: \dfrac{{sec}^{2}x }{ - 6tanx - 4 + 4 {tan}^{2} x} \: dx$

$\rm  \:  =  \: - \:\displaystyle\int\sf \: \dfrac{{sec}^{2}x }{4 {tan}^{2} x - 6tanx - 4} \: dx$

$\red{\rm :\longmapsto\:Put \: tanx = y} \\ \red{\rm :\longmapsto\: {sec}^{2}xdx = dy}$

$\rm  \:  =  \: \: - \: \displaystyle\int\sf \: \dfrac{dy}{4 {y}^{2} - 6y - 4}$

$\rm  \:  =  \: \: - \dfrac{1}{2} \: \displaystyle\int\sf \: \dfrac{dy}{2 {y}^{2} - 3y - 2}$

$\rm  \:  =  \: \: - \dfrac{1}{2} \: \displaystyle\int\sf \: \dfrac{dy}{2 {y}^{2} - 4y + y- 2}$

$\rm  \:  =  \: \: - \dfrac{1}{2} \: \displaystyle\int\sf \: \dfrac{dy}{2y(y - 2) +1( y- 2)}$

$\rm  \:  =  \: \: - \dfrac{1}{2} \: \displaystyle\int\sf \: \dfrac{dy}{(2y + 1)(y - 2)}$

$\rm  \:  =  \: \: - \dfrac{1}{10} \: \displaystyle\int\sf \: \dfrac{5}{(2y + 1)(y - 2)} dy$

$\rm  \:  =  \: \: - \dfrac{1}{10} \: \displaystyle\int\sf \: \dfrac{4 + 1}{(2y + 1)(y - 2)} dy$

$\rm  \:  =  \: \: - \dfrac{1}{10} \: \displaystyle\int\sf \: \dfrac{4 + 1 + 2y - 2y}{(2y + 1)(y - 2)} dy$

$\rm  \:  =  \: \: - \dfrac{1}{10} \: \displaystyle\int\sf \: \dfrac{(2y + 1) - 2(y - 2)}{(2y + 1)(y - 2)} dy$

$\rm  \:  =  \: \: - \dfrac{1}{10} \: \displaystyle\int\sf \: \bigg(\dfrac{1}{y - 2} - \dfrac{2}{2y + 1} \bigg) dy$

$\rm  \:  =  \: \: - \dfrac{1}{10}\bigg( log(y - 2) - log(2y + 1) \bigg) + c$

$\rm  \:  =  \: \: - \dfrac{1}{10}\bigg( log(tanx - 2) - log(2tanx + 1) \bigg) + c$

Formula used :-

$\green{ \boxed{ \bf \: \displaystyle\int\sf \: \frac{1}{x}dx = log(x) + c}}$

$\green{ \boxed{ \bf \: \displaystyle\int\sf \: \frac{1}{ax + b} \: dx = \frac{1}{a} \: log(ax + b) + c}}$

Additional Information :-

$\green{ \boxed{ \bf \: \displaystyle\int\sf \: k \: dx = kx \: + \: c}}$

$\green{ \boxed{ \bf \: \displaystyle\int\sf \: sinx \: dx = - \: cosx \: + \: c}}$

$\green{ \boxed{ \bf \: \displaystyle\int\sf \: cosx \: dx = \: sinx \: + \: c}}$

$\green{ \boxed{ \bf \: \displaystyle\int\sf \: cosecx \: cotx \: dx = \: - \: cosecx \: \: + \: c}}$

$\green{ \boxed{ \bf \: \displaystyle\int\sf \: secx \: tanx \: dx = \: secx \: \: + \: c}}$

$\green{ \boxed{ \bf \: \displaystyle\int\sf \: secx\: dx = \: log(secx + tanx) \: \: + \: c}}$

$\green{ \boxed{ \bf \: \displaystyle\int\sf \: cosecx\: dx = \: log(cosecx - cotx) \: \: + \: c}}$

$\green{ \boxed{ \bf \: \displaystyle\int\sf \: {sec}^{2}x = tanx \: + \: c}}$

$\green{ \boxed{ \bf \: \displaystyle\int\sf \: {cosec}^{2}x = - \: cotx \: + \: c}}$