Question

17. Evaluate: integrate (root tan x - root cotx) dx

Answer 1

$\rm \longrightarrow \displaystyle \int \rm \:( \sqrt{tan \: x} \: \: - \sqrt{cot \: x} ) dx$

$\rm \longrightarrow \displaystyle \int \rm \bigg( \frac{ \sqrt{sin \: x} }{ \sqrt{cos \: x} } \: \: -\frac{ \sqrt{ cos\: x} }{ \sqrt{sin \: x} } \bigg ) dx$

$\rm \longrightarrow \displaystyle \int \rm \bigg( \dfrac{ {sin \: x} - cos \: x}{ \sqrt{sin \: x \: cos \: x} } \: \bigg) dx$

$\rm \longrightarrow \displaystyle \int \rm \dfrac{ \sqrt{2} ({sin \: x} - cos \: x)}{ \sqrt{2 \: sin \: x \: cos \: x} } \: dx$

Now , 2 sinx cosx = ( sinx + cosx )² - 1

$\rm \longrightarrow \displaystyle \int \rm \dfrac{ \sqrt{2} ({sin \: x} - cos \: x)}{ \sqrt{ {(sin \: x + cos \: x)}^{2} - 1} } \: dx$

$\rm \longrightarrow \sqrt{2}\displaystyle \int \rm \dfrac{ ({sin \: x} - cos \: x)}{ \sqrt{ {(sin \: x + cos \: x)}^{2} - 1} } \: dx$

Let , sinx + cosx = t

➡️(cosx - sinx ) dx = dt

➡️(- cosx + sinx ) dx = - dt

$\rm \longrightarrow \sqrt{2}\displaystyle \int \rm \dfrac{ - dt}{ \sqrt{ {t}^{2} - 1} }$

$\rm \longrightarrow \sqrt{2}\displaystyle \int \rm \dfrac{ - dt}{ \sqrt{ {t}^{2} - 1} } = - \: \sqrt{2} \bigg( log \bigg|t + \sqrt{ {t}^{2} - 1 }\bigg| \bigg) + c$

Now , put t = sinx + cosx

$\rm \longrightarrow - \: \sqrt{2} \: log \bigg|sin \: x + cos \: x + \sqrt{ {(sin \: x + cos \: x)}^{2} - 1 }\bigg| + c$

$\rm \longrightarrow - \: \sqrt{2} \: log \bigg|sin \: x + cos \: x + \sqrt{ 2 \: cos \: \: sin \: x }\bigg| + c$