Question

Evaluate : Integral dx
/4+5sinx

Answer 1

Step-by-step explanation:

this is the way to solve this

Answer 2

$\rm :\longmapsto\:\rm \displaystyle \int\dfrac{dx}{4 + 5sinx} - - - (1)$

We know,

$\rm :\longmapsto\: \boxed{ \bf{ \: sinx = \dfrac{2tan\dfrac{x}{2} }{1 + {tan}^{2}\dfrac{x}{2} } }}$

• So, equation (1) reduces to

$\rm :\longmapsto\:\rm \displaystyle \int \sf \: \dfrac{dx}{4 + 5 \times \dfrac{2tan\dfrac{x}{2}}{1 + {tan}^{2}\dfrac{x}{2} } }$

$\rm :\longmapsto\:\rm \displaystyle \int \sf \: \dfrac{dx}{4 + \dfrac{10tan\dfrac{x}{2}}{1 + {tan}^{2}\dfrac{x}{2} } }$

$= \rm \displaystyle \int \: \sf \: \dfrac{1 + {tan}^{2} \dfrac{x}{2}}{4 + 4 {tan}^{2}\dfrac{x}{2} + 10tan\dfrac{x}{2} } dx$

$= \rm \displaystyle \int \: \sf \: \dfrac{{sec}^{2} \dfrac{x}{2}}{4 + 4 {tan}^{2}\dfrac{x}{2} + 10tan\dfrac{x}{2} } dx$

As we know that,

• By using substitution method, we get.

$\rm :\longmapsto\:Put \: tan\dfrac{x}{2} = y$

• Differentiate w.r.t x, we get.

$\rm :\longmapsto\: {sec}^{2} \dfrac{x}{2} \times \dfrac{1}{2} = \dfrac{dy}{dx}$

$\rm :\longmapsto\: {sec}^{2}\dfrac{x}{2}dx = 2dy$

On substituting all these values in above integral, we get

$= \rm \displaystyle \int \: \sf \: \dfrac{2dy}{4 {y}^{2} + 10y + 4}$

$= \rm \displaystyle \int \: \sf \: \dfrac{dy}{ {2y}^{2} + 5y + 2}$

$= \rm \displaystyle \dfrac{1}{2} \int \: \sf \: \dfrac{dy}{ {y}^{2} + \dfrac{5}{2} y + 1 }$

$= \rm \displaystyle \dfrac{1}{2} \int \: \sf \: \dfrac{dy}{ {y}^{2} + \dfrac{5}{2} y + \dfrac{25}{16} - \dfrac{25}{16} + 1 }$

$= \rm \displaystyle \dfrac{1}{2} \int \: \sf \: \dfrac{dy}{ {\bigg(y + \dfrac{5}{4} \bigg) }^{2} - \dfrac{9}{16} }$

$= \rm \displaystyle \dfrac{1}{2} \int \: \sf \: \dfrac{dy}{ {\bigg(y + \dfrac{5}{4} \bigg) }^{2} - {\bigg( \dfrac{3}{4} \bigg) }^{2} }$

$= \sf \: \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{4}{3} log\bigg(\dfrac{y + \dfrac{5}{4} - \dfrac{3}{4} }{y + \dfrac{5}{4} + \dfrac{3}{4} } \bigg) + c$

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

$= \sf \: \dfrac{1}{3} \: log\bigg( \dfrac{y + \dfrac{1}{2} }{y + 2} \bigg) + c$

$= \sf \: \dfrac{1}{3} \: log\bigg(\dfrac{2y + 1}{2(y + 2)} \bigg) + c$

$= \sf \: \dfrac{1}{3} \: log\bigg(\dfrac{2y + 1}{(y + 2)} \bigg) + \dfrac{1}{3} log\dfrac{1}{2} + c$

$\sf \: = \sf \: \dfrac{1}{3} \: log\bigg(\dfrac{2y + 1}{y + 2} \bigg) + d$

$\sf \: = \sf \: \dfrac{1}{3} \: log\bigg(\dfrac{2tan\dfrac{x}{2} + 1}{tan\dfrac{x}{2} + 2} \bigg) + d$

$\: \: \: \: \: \: \boxed{ \bf{ \: on \: substituting \: the \: value \: of \:y = tan\dfrac{x}{2}}}$

More information :-

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

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