Question

∫ ( sin^2 x cos^2 x / sin x cos x) dx

Answer 1

$\huge{ \underline{ \underline{ \mathfrak{ \blue{answer \: : - }}}}}$

$\huge \sf \longmapsto I \: = \int \frac{ {sin}^{2}x \: - {cos}^{2}x }{sin \: x \: - cos \: x} dx \\ \\ \huge \sf \longmapsto I \: = \int \frac{(sin \: x \: + \: cos \: x)(sin \: x - cos \: x)}{sin \: x \: - cos \: x} \\ \\ \huge \sf \longmapsto I \: = \int(sin \: x \: + cos \: x)dx \\ \\ \huge \sf \longmapsto I \: = \int \: sin \: xdx \: + \int \: cos \: xdx \\ \\ \huge \sf \longmapsto I \: = - cos \: x \: + sin \: x + c \\ \\ \huge \sf \longmapsto I \: = sin \: x \: - cos \: x \: + c...$

Answer 2

Answer

sin²x/2 + c

$\bf\large\underline{Solution}$

$\rm\longrightarrow \int \dfrac{ \sin ^{2} x \cos {}^{2}x }{ \sin x\cos x } dx\\ \\ \rm\longrightarrow \int \dfrac{(\sin x \cos {x})^{2} }{ \sin x \cos x } dx\\ \\ \rm\longrightarrow \int \sin x \cos x dx$

Let us consider ,

$\rm\implies t = \sin x \\\\ \rm\implies dt = d(\sin x) \\\\ \rm\implies dt = \cos x dx \dashrightarrow (1)$

Thus we have ,

$\rm\longrightarrow \int \sin x \cos xdx \\ \\ \rm\longrightarrow \int tdt \\ \\ \rm\longrightarrow \dfrac{ {t}^{2} }{2} + c \\ \\ \rm\longrightarrow \dfrac{ \sin {}^{2} x }{2} + c$