Question

Intergrate
$\to \bf \int \dfrac{dx}{sin {}^{2} xcos^{2}x }$
Option are
A) tanx - cotx + k
B) tanx + cotx + k
C) - tanx + cotx + k
D) - tanx - cotx + k ​

Answer 1

$\rm \longrightarrow \displaystyle \int \rm\dfrac{dx}{sin {}^{2} x \: cos^{2}x }$

$\rm \longrightarrow \displaystyle \int \rm\dfrac{1}{sin {}^{2} x \: cos^{2}x } \: dx$

We know that :-

$\boxed{ \rm \large {sin}^{2} \theta + {cos}^{2} \theta= 1}$

$\rm \longrightarrow \displaystyle \int \rm\dfrac{{sin}^{2} x + {cos}^{2} x}{sin {}^{2} x \: cos^{2}x } \: dx$

$\rm \longrightarrow \displaystyle \int \bigg(\rm\dfrac{{sin}^{2} x }{sin {}^{2} x \: cos^{2}x } + \frac{{cos}^{2} x}{sin {}^{2} x \: cos^{2}x} \bigg) dx$

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

$\rm \longrightarrow \displaystyle \int \bigg( \rm \: {sec}^{2}x + {cosec}^{2} x \bigg) dx$

$\rm \longrightarrow \displaystyle \int \rm \: {sec}^{2}x \: dx + \int {cosec}^{2} x \: dx$

$\rm \longrightarrow \displaystyle \rm tan \: x - cot \: x + k$

Therefore, option A) tanx - cotx + k is correct

Answer 2

Question:-

$\to\bf\int \dfrac{dx}{ \sin^{2}x \: { \cos }^{2}x}$

Given Options:-

• A) tanx - cotx + k
• B) tanx + cotx + k
• C) - tanx + cotx + k
• D) - tanx - cotx + k

Solution:-

• Option A) tanx - cotx + k

Explanation:-

$\to \: \bf\int \frac{dx}{ {sin}^{2}x \: {cos}^{2} x}$

$\to\bf \int \dfrac{1}{ {sin}^{2}x \: {cos}^{2}x}dx$

$\sf\green{as \: we \: know \: that : }$

$\rm\large {sin}^{2}\theta + {cos}^{2}\theta = 1$

So,

$\implies \: \bf\int \: \dfrac{ {sin}^{2}x \: + {cos}^{2}x }{ {sin}^{2}x \: {cos}^{2}x } dx$

$\implies \: \bf\int( \dfrac{ {sin}^{2}x}{ {sin}^{2}x \: {cos}^{2}x} + \dfrac{ {cos}^{2}x }{ {sin}^{2}x \: {cos}^{2}x } dx)$

$\implies \: \bf\int( \dfrac{1}{ {cos}^{2}x} + \dfrac{1}{ {sin}^{2}x} )dx$

$\implies \: \bf\int \: ({sec}^{2} + {cosec}^{2}x) \: dx$

$\implies \: \bf\int \: {sec}^{2}x \: dx + \: \int {cosec}^{2}dx$

$\implies \: \bf \: tan \: x - cot \: x + k$

Hence, Option A is correct :)