Question

What the integral of 1÷(1-sinxcosx)

Answer 1

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

$\rm :\longmapsto\: \displaystyle\rm \: \: \: \int \: \dfrac{dx}{1 - sinx \: cosx}$

$\red{ \tt \: Divide \: numerator \: and \: denominator \: by \: {cos}^{2}x}$

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

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

$\: \: \: \: \: \: \: \: \: \: \: \: \green{\boxed{ \bf{ \: {sec}^{2}x = 1 + {tan}^{2}x}}}$

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

Now, we use method of Substitution

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

So, on substituting the values, we get

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

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

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

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

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

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

$\rm \: \: = \: \dfrac{1}{\dfrac{ \sqrt{3} }{2} } {tan}^{ - 1}\bigg(\dfrac{\dfrac{y - \dfrac{1}{2} }{ \sqrt{3} } }{2} \bigg) + c$

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

$\rm \: \: = \: \dfrac{2}{ \sqrt{3} } {tan}^{ - 1}\bigg(\dfrac{2y - 1}{ \sqrt{3} } \bigg) + c$

$\rm \: \: = \: \dfrac{2}{ \sqrt{3} } {tan}^{ - 1}\bigg(\dfrac{2tanx - 1}{ \sqrt{3} } \bigg) + c$

Additional Information :-

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

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

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

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