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

1 thought on “integration of 1/(3sin2x+4cos2x)<br />note: 1/(3 sin2x+ 4 cos 3x) it is .there is no square​​”

1. [tex]\large\underline{\sf{Solution-}}[/tex]

   [tex]\rm :\longmapsto\:\displaystyle\int\sf \: \dfrac{1}{3sin2x \: + \: 4cos2x} \: dx[/tex]

   We know, that

   [tex]\green{ \boxed{ \bf \: sin2x = \frac{2tanx}{1 + {tan}^{2}x }}}[/tex]

   and

   [tex]\green{ \boxed{ \bf \: cos2x = \frac{1 – {tan}^{2}x }{1 + {tan}^{2}x }}}[/tex]

   So, on substituting these Identities, we get

   [tex]\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[/tex]

   [tex]\rm \: = \: \:\displaystyle\int\sf \: \dfrac{1 + {tan}^{2}x }{6tanx + 4 – 4 {tan}^{2} x} \: dx[/tex]

   [tex]\rm \: = \: – \:\displaystyle\int\sf \: \dfrac{{sec}^{2}x }{ – 6tanx – 4 + 4 {tan}^{2} x} \: dx[/tex]

   [tex]\rm \: = \: – \:\displaystyle\int\sf \: \dfrac{{sec}^{2}x }{4 {tan}^{2} x – 6tanx – 4} \: dx[/tex]

   [tex] \red{\rm :\longmapsto\:Put \: tanx = y} \\ \red{\rm :\longmapsto\: {sec}^{2}xdx = dy}[/tex]

   [tex]\rm \: = \: \: – \: \displaystyle\int\sf \: \dfrac{dy}{4 {y}^{2} – 6y – 4} [/tex]

   [tex]\rm \: = \: \: – \dfrac{1}{2} \: \displaystyle\int\sf \: \dfrac{dy}{2 {y}^{2} – 3y – 2} [/tex]

   [tex]\rm \: = \: \: – \dfrac{1}{2} \: \displaystyle\int\sf \: \dfrac{dy}{2 {y}^{2} – 4y + y- 2} [/tex]

   [tex]\rm \: = \: \: – \dfrac{1}{2} \: \displaystyle\int\sf \: \dfrac{dy}{2y(y – 2) +1( y- 2)} [/tex]

   [tex]\rm \: = \: \: – \dfrac{1}{2} \: \displaystyle\int\sf \: \dfrac{dy}{(2y + 1)(y – 2)} [/tex]

   [tex]\rm \: = \: \: – \dfrac{1}{10} \: \displaystyle\int\sf \: \dfrac{5}{(2y + 1)(y – 2)} dy[/tex]

   [tex]\rm \: = \: \: – \dfrac{1}{10} \: \displaystyle\int\sf \: \dfrac{4 + 1}{(2y + 1)(y – 2)} dy[/tex]

   [tex]\rm \: = \: \: – \dfrac{1}{10} \: \displaystyle\int\sf \: \dfrac{4 + 1 + 2y – 2y}{(2y + 1)(y – 2)} dy[/tex]

   [tex]\rm \: = \: \: – \dfrac{1}{10} \: \displaystyle\int\sf \: \dfrac{(2y + 1) – 2(y – 2)}{(2y + 1)(y – 2)} dy[/tex]

   [tex]\rm \: = \: \: – \dfrac{1}{10} \: \displaystyle\int\sf \: \bigg(\dfrac{1}{y – 2} – \dfrac{2}{2y + 1} \bigg) dy[/tex]

   [tex]\rm \: = \: \: – \dfrac{1}{10}\bigg( log(y – 2) – log(2y + 1) \bigg) + c[/tex]

   [tex]\rm \: = \: \: – \dfrac{1}{10}\bigg( log(tanx – 2) – log(2tanx + 1) \bigg) + c[/tex]

   Formula used :-

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

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

   Additional Information :-

   [tex]\green{ \boxed{ \bf \: \displaystyle\int\sf \: k \: dx = kx \: + \: c}}[/tex]

   [tex]\green{ \boxed{ \bf \: \displaystyle\int\sf \: sinx \: dx = – \: cosx \: + \: c}}[/tex]

   [tex]\green{ \boxed{ \bf \: \displaystyle\int\sf \: cosx \: dx = \: sinx \: + \: c}}[/tex]

   [tex]\green{ \boxed{ \bf \: \displaystyle\int\sf \: cosecx \: cotx \: dx = \: – \: cosecx \: \: + \: c}}[/tex]

   [tex]\green{ \boxed{ \bf \: \displaystyle\int\sf \: secx \: tanx \: dx = \: secx \: \: + \: c}}[/tex]

   [tex]\green{ \boxed{ \bf \: \displaystyle\int\sf \: secx\: dx = \: log(secx + tanx) \: \: + \: c}}[/tex]

   [tex]\green{ \boxed{ \bf \: \displaystyle\int\sf \: cosecx\: dx = \: log(cosecx – cotx) \: \: + \: c}}[/tex]

   [tex]\green{ \boxed{ \bf \: \displaystyle\int\sf \: {sec}^{2}x = tanx \: + \: c}} [/tex]

   [tex]\green{ \boxed{ \bf \: \displaystyle\int\sf \: {cosec}^{2}x = – \: cotx \: + \: c}} [/tex]

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