Question

solve the above question​

Answer 1

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

$\rm :\longmapsto\:\displaystyle\int\sf \dfrac{ {cos}^{2} x}{ {sin}^{4} x} dx$

can be rewritten as

$\rm \: \: = \: \:\displaystyle\int\sf \dfrac{ {cos}^{2} x}{ {sin}^{2} \: x \: \times {sin}^{2} x} dx$

$\rm \: \: = \: \displaystyle\int\sf \bigg(\dfrac{ {cos}^{2}x }{ {sin}^{2} x} \times \dfrac{1}{ {sin}^{2}x } \bigg)$

$\rm \: \: = \: \displaystyle\int\sf {cot}^{2}x \: {cosec}^{2}x \: dx$

Now, we use method of Substitution,

$\red{\rm :\longmapsto\:Let \: cotx \: = \: y}$

On differentiating both sides w. r. t. x, we get

$\red{\rm :\longmapsto\:\dfrac{d}{dx} \: cotx \: = \:\dfrac{d}{dx} y}$

$\red{\rm :\longmapsto\: - \: {cosec}^{2}x \: = \dfrac{dy}{dx}}$

$\red{\rm :\longmapsto\: \: {cosec}^{2}x \: dx\: = - \: dy}$

So, on substituting all these values, we get

$\rm \: \: = \: - \: \displaystyle\int\sf {y}^{2}dy$

$\rm \: \: = \: - \: \dfrac{ {y}^{2 + 1} }{2 + 1} + c$

$\rm \: \: = \: - \: \dfrac{ {y}^{3} }{3} + c$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \boxed{ \sf{ \because \: \: \displaystyle\int\sf {x}^{n} dx = \dfrac{ {x}^{n + 1} }{n + 1} + c}}}$

$\rm \: \: = \: - \: \dfrac{ {cot}^{3}x}{3} + c$

Hence,

$\bf\implies \:\displaystyle\int\sf \dfrac{ {cos}^{2} x}{ {sin}^{4} x} dx \rm \: \: = \: - \: \dfrac{ {cot}^{3}x}{3} + c$

Additional Information :-

$\red{ \boxed{ \sf{ \:\displaystyle\int\sf k \: dx = kx \: + \: c }}}$

$\red{ \boxed{ \sf{ \:\displaystyle\int\sf {e}^{x} \: dx = {e}^{x} \: + \: c }}}$

$\red{ \boxed{ \sf{ \:\displaystyle\int\sf {a}^{x} \: dx = \frac{ {a}^{x} }{loga} \: + \: c \: \: where \: a > 0 }}}$

$\red{ \boxed{ \sf{ \:\displaystyle\int\sf cosx \: dx = \: sinx \: + \: c }}}$

$\red{ \boxed{ \sf{ \:\displaystyle\int\sf sinx \: dx = \: - \: cosx \: + \: c }}}$

$\red{ \boxed{ \sf{ \:\displaystyle\int\sf secx \: tanx\: dx = \: \: secx \: + \: c }}}$

$\red{ \boxed{ \sf{ \:\displaystyle\int\sf cosecx \: cotx\: dx = \: - \: \: cosecx \: + \: c }}}$

$\red{ \boxed{ \sf{ \:\displaystyle\int\sf tanx \: = \: log \: secx \: + \: c }}}$

$\red{ \boxed{ \sf{ \:\displaystyle\int\sf cotx \: = \: log \: sinx \: + \: c }}}$

$\red{ \boxed{ \sf{ \:\displaystyle\int\sf cosecx \: = \: log \: (cosecx + cotx)\: + \: c }}}$

$\red{ \boxed{ \sf{ \:\displaystyle\int\sf secx \: = \: log \: (secx + tanx)\: + \: c }}}$