Question

$\large \displaystyle \sf \pink{\int \bigg( \sqrt{ cotx } - \sqrt{tanx} \bigg ) \: dx}$​

Answer 1

Answer:

$\purple{ \frac{1}{2 \sqrt{2} } \log \bigg| \frac{ \sqrt{ \tan(x) } + \sqrt{ \cot(x) } + \sqrt{2} }{ \sqrt{ \tan(x) } + \sqrt{ \cot(x) } - \sqrt{2} } \bigg|+C }$

Answer 2

STEP-BY-STEP EXPLANATION:

.

$A = \displaystyle \sf {\int \bigg( \sqrt{ cotx } - \sqrt{tanx} \bigg ) \: dx}$

$Let's \: Substitute,$

$\implies \sf \tan x = {y}^{2}$

$\implies \sf { \sec}^{2} x = 1 + \tan^{2} x$

$\implies \sf { \sec}^{2} x = 1 + y^{4}$

$\implies \sf { \sec} ^{2} x \: \sf{dx} = 2 y \: dy$

$\implies \sf \sf{dx} = \frac{2 y }{ { \sec }^{2} x} \: dy$

$\implies \sf \sf{dx} = \frac{2 y }{ 1 + {y}^{4} } \: dy$

$As \: we \: know \: that,$

• $\sf \cot A = \frac{1}{ \tan A }$

$\implies A = \displaystyle \sf {\int \bigg( \sqrt{ cotx } - \sqrt{tanx} \bigg ) \: dx}$

$\implies A = \displaystyle \sf {\int \bigg( \sqrt{ \frac{1}{ {y}^{2} } } - \sqrt{ {y}^{2} } \bigg ) \frac{2y}{1 + {y}^{4} } \: dy }$

$\implies A = \displaystyle \sf {\int \bigg( \frac{1}{y} - y \bigg ) \frac{2y}{1 + {y}^{4} } \: dy }$

$\implies A = \displaystyle \sf {\int \bigg( \frac {1 - {y}^{2} }{y} \bigg ) \frac{2y}{1 + {y}^{4} } \: dy }$

$\implies A = \displaystyle \sf {\int \frac{1 - {y}^{2} }{1 + {y}^{4} } \: dy \times \frac{ \frac{1}{ {y}^{2} } }{ \frac{1}{ {y}^{2} } } }$

$\implies A = \displaystyle \sf {\int \frac{ \frac{1}{ {y}^{2} } - 1 }{ {y}^{2} + \frac{1}{ {y}^{2} } } \: dy }$

$\implies A = \displaystyle \sf {\int \frac{ \frac{1}{ {y}^{2} } - 1 }{ {y}^{2} + \frac{1}{ {y}^{2} } + 2 - 2 } \: dy }$

$\implies A = \displaystyle \sf {\int \frac{ \frac{1}{ {y}^{2} } - 1 }{ {(y + \frac{1}{y}) }^{2} - 2 } \: dy } \: \: \: \: \: \: \: \: \: \: \because {a}^{2} + {b}^{2} + 2ab = (a + b) ^{2}$

$Let's \: Substitute,$

• $\sf y + \frac{1}{y} = f$

$\sf ( \frac{1}{ {y}^{2} } - 1)dy = df$

$We \: find,$

$\implies A = \displaystyle \sf {\int \frac{ df }{ {f}^{2} - 2 } }$

$\implies A = \displaystyle \sf {\int \frac{ df }{ {f}^{2} - 2 } } = \frac{1}{2 \sqrt{2} } log \bigg( \frac{f + \sqrt{2} }{f - \sqrt{2} } \bigg) + c$

$As \: we \: know \: that,$

$\implies \sf y + \frac{1}{y} = f$

$\implies \sf \sqrt{tanx} + \sqrt{cotx} = f$

$Therefore,$

$\implies\sf A = \frac{1}{2 \sqrt{2} } log \bigg( \frac{f + \sqrt{2} }{f - \sqrt{2} } \bigg) + c$

$\implies \sf A = \frac{1}{2 \sqrt{2} } log \bigg( \frac{ \sqrt{tanx} + \sqrt{cotx} + \sqrt{2} }{ \sqrt{tanx} + \sqrt{cotx} - \sqrt{2} } \bigg) + c$

REQUIRED ANSWER,

.

• $\sf A = \frac{1}{2 \sqrt{2} } log \bigg( \frac{ \sqrt{tanx} + \sqrt{cotx} + \sqrt{2} }{ \sqrt{tanx} + \sqrt{cotx} - \sqrt{2} } \bigg) + c$