Question

Evaluate :dx/ 3-4 sin2x​

Answer 1

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

$\rm :\longmapsto\:\displaystyle\int \sf \:\dfrac{dx}{3 - 4 \: sin2x}$

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

$\rm \: \: = \: \displaystyle\int \sf \:\dfrac{dx}{ \: \dfrac{3 + 3 {tan}^{2}x - 8tanx}{1 + {tan}^{2}x }}$

$\rm \: \: = \: \displaystyle\int \sf \:\dfrac{1 + {tan}^{2}x}{3 {tan}^{2}x + 3 - 8tanx } dx$

$\rm \: \: = \: \displaystyle\int \sf \:\dfrac{{sec}^{2}x}{3 {tan}^{2}x + 3 - 8tanx } dx$

Now we use substitution method to solve the above integral

$\red{\rm :\longmapsto\:Let \: tanx = y} \\ \red{\rm :\longmapsto\: {sec}^{2}x \: dx = dy}$

$\rm \: \: = \: \displaystyle\int \sf \:\dfrac{dy}{3 {y}^{2} - 8y + 3}$

$\rm \: \: = \: \dfrac{1}{3} \displaystyle\int \sf \:\dfrac{dy}{ {y}^{2} - \dfrac{8}{3} y + 1}$

Using completing squares method

$\rm \: \: = \: \dfrac{1}{3} \displaystyle\int \sf \:\dfrac{dy}{ {y}^{2} - \dfrac{8}{3} y + \dfrac{16}{9} - \dfrac{16}{9} + 1}$

$\rm \: \: = \: \dfrac{1}{3} \displaystyle\int \sf \:\dfrac{dy}{ {\bigg(y - \dfrac{4}{3} \bigg) }^{2} + \dfrac{9 - 16}{9}}$

$\rm \: \: = \: \dfrac{1}{3} \displaystyle\int \sf \:\dfrac{dy}{ {\bigg(y - \dfrac{4}{3} \bigg) }^{2} - \dfrac{7}{9}}$

$\rm \: \: = \: \dfrac{1}{3} \displaystyle\int \sf \:\dfrac{dy}{ {\bigg(y - \dfrac{4}{3} \bigg) }^{2} - {\bigg(\dfrac{ \sqrt{7} }{3} \bigg) }^{2} }$

$\rm \: \: = \: \dfrac{1}{3} \times \dfrac{1}{2 \times \dfrac{ \sqrt{7} }{3} }log\bigg(\dfrac{y - \dfrac{4}{3} - \dfrac{ \sqrt{7} }{3} }{y - \dfrac{4}{3} + \dfrac{ \sqrt{7} }{3} } \bigg) + c$

$\: \: \: \: \: \red{\bigg \{ \because \: \displaystyle\int \sf \:\dfrac{dx}{ {x}^{2} - {a}^{2} } = \dfrac{1}{2a}log\bigg(\dfrac{x - a}{x + a}\bigg) + c \bigg \}}$

$\rm \: \: = \: \dfrac{1}{2 \sqrt{7} } log\bigg(\dfrac{3y - 4 - \sqrt{7} }{3y - 4 + \sqrt{7} } \bigg) + c$

$\rm \: \: = \: \dfrac{1}{2 \sqrt{7} } log\bigg(\dfrac{3tanx - 4 - \sqrt{7} }{3tanx - 4 + \sqrt{7} } \bigg) + c$

Additional Information :-

$\green{\boxed{ \tt \:\displaystyle\int \sf \:\dfrac{dx}{ {x}^{2} + {a}^{2} } = \dfrac{1}{a} {tan}^{ - 1} \dfrac{x}{a} + c}}$

$\green{\boxed{ \tt \:\displaystyle\int \sf \:\dfrac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } } = {sin}^{ - 1}\dfrac{x}{a} + c}}$

$\green{\boxed{ \tt \:\displaystyle\int \sf \:\dfrac{dx}{ \sqrt{ {x}^{2} - {a}^{2} } } = log |x + \sqrt{ {x}^{2} - {a}^{2} } | + c}}$

$\green{\boxed{ \tt \:\displaystyle\int \sf \:\dfrac{dx}{ \sqrt{ {x}^{2} + {a}^{2} } } = log |x + \sqrt{ {x}^{2} + {a}^{2} } | + c}}$