Question

Please answer just now ​

Answer 1

Step-by-step explanation:

Answer and solution given in above picture

Answer 2

Question given:

• $\sf{ \large{\displaystyle\int{}}\dfrac{dx}{sin {}^{2} xcos {}^{2} x} }$

Solution:

Here I is calculas,

• $: \longmapsto \: \sf{ I \: = \: \large{\displaystyle\int{}}\dfrac{dx}{sin {}^{2} xcos {}^{2} x} }$

Putting (sec²)²x in numerator,

$: \longmapsto \: \sf{ I \: = \: \large{\displaystyle\int{}}\dfrac{(sec {}^{2}) {}^{2} x }{sin {}^{2} xcos {}^{2} x} \: dx }$

Dividing cos⁴ in numerator,

$: \longmapsto \: \sf{ I \: = \: \large{\displaystyle\int{}}\dfrac{(sec {}^{2}) {}^{2} x }{sin {}^{2} xcos {}^{2} x \: \div \: cos {}^{4} } \: dx }$

$: \longmapsto \: \sf{ I \: = \: \large{\displaystyle\int{}}\dfrac{(sec {}^{2}) {}^{2} x }{sin {}^{2} xcos {}^{2} x \times \dfrac{1}{cos {}^{4} } } \: dx }$

$:\longmapsto \: \sf{ I \: = \: \large{\displaystyle\int{}}\dfrac{(sec {}^{2}) {}^{2} x }{sin {}^{2} x \cancel{cos {}^{2}} x \times \dfrac{1}{ \cancel{cos {}^{4} }} } \: dx }$

$:\longmapsto \: \sf{ I \: = \: \large{\displaystyle\int{}}\dfrac{(sec {}^{2}) {}^{2} x }{sin {}^{2} x \: \times \: \dfrac{1}{cos {}^{2} } } \: dx }$

$:\longmapsto \: \sf{ I \: = \: \large{\displaystyle\int{}}\dfrac{(sec {}^{2}) {}^{2} x }{ \dfrac{sin {}^{2}x }{cos {}^{2}x } } \: dx }$

We know that,

• $\boxed{ \sf{ \dfrac{sin {}^{2} }{cos {}^{2} } \: = \: tan {}^{2} x }}$

By using it we gets,

$:\longmapsto \: \sf{ I \: = \: \large{\displaystyle\int{}}\dfrac{(sec {}^{2}) }{tan {}^{2}x } \: dx }$

$:\longmapsto \: \sf{ I \: = \: \large{\displaystyle\int{}}\dfrac{(sec {}^{2} ) {}^{2}x }{tan {}^{2}x } \: dx }$

$:\longmapsto \: \sf{ I \: = \: \large{\displaystyle\int{}}\dfrac{sec {}^{2} x \times sec {}^{2}x }{tan {}^{2}x } \: dx }$

Using,

• $\boxed{ \sf{sec {}^{2}x \: = \: 1 + tan {}^{2} x}}$

Substituting the value of sec²x as 1 + tan²x,

$:\longmapsto \: \sf{ I \: = \: \large{ \displaystyle\int{}}\dfrac{(1 + tan {}^{2} x) sec {}^{2}x }{tan {}^{2}x } \: dx }$

• Also assuming that tanx as t , sec²x dx as dt and cot x as 1/2. It would ease our calculations. tan²x would become t² because tan²x is in square and substituting them,

$:\longmapsto \: \sf{ I \: = \: \large{ \displaystyle\int{}}\dfrac{(1 + t {}^{2}) dt}{t{}^{2}} \: }$

Now seperating them and forming two different fractions,

$:\longmapsto \: \sf{ I \: = \: \large{ \displaystyle\int{}}dt \: + \: \displaystyle \int{} \dfrac{dt}{t {}^{2} } }$

Now it would become,

$:\longmapsto \: \sf{ I \: = \: \large{t \: - \: \dfrac{1}{t } \: + \: C \: } }$

As we have assumed,

• tan x as t
• 1/t as cotx

Substituting them,

$:\longmapsto \: \underline{ \boxed{\sf{ I \: = \: \large{tan \: x\: - \: cot \: x \: + \: C \: } }}}$

$\therefore \underline{\bf{tan \: x\: - \: cot \: x \: + \: C \: is \: the \: required \: answer}}$