Question

Given, \int\left\lbrack \dfrac{\text{cotxdx}}{\left\\{\sqrt\ (cos^{4}x+\ sin^{4}x) \ \right\\}} \right\rbrack = \ \\_\\_\\_\\_\\_\\_\\_\\_\\_\ + \ c\ \
(a)$\ \left( \dfrac{1}{2} \right)\log\left| cot^{2}x\ + \ \ {\sqrt\ (cot^{4} + 1}) \right|$
(b) $\ \left(- \dfrac{1}{2} \right)\log\left| cot^{2}x\ + \ \ {\sqrt\ (cot^{4} + 1}) \right|$
(c) $\left( \dfrac{1}{2} \right)\log\left| tan^{2}x\ + \ \ {\sqrt\ (tan^{4} + 1}) \right|$
(d) $\ \left( \dfrac{1}{2} \right)\log\left| cotx\ + \ \ {\sqrt\ (cot^{4} + 1}) \right|$

Answer 1

In this question, we need to integrate the given expression. Here we will use the basic trigonometric identities to solve this question . The symbol `$\int$’ is the sign of integration. The process of finding the integral is known as integration.
Formula used :
$\dfrac{1}{\sqrt\ {(x^{2}+a^{2})}}{dx}=\dfrac{1}{2}|log\ \left( x +{\sqrt\ {(x^{2}+a^{2})}}{dx} \ \right)|\ \ + c$

Complete answer: Given,
\int\left\lbrack \dfrac{\text{cotxdx}}{\left\\{\sqrt\ (cos^{4}x+\ sin^{4}x)\right\\}} \right\rbrack
Let us consider this as I=\int\left\lbrack \dfrac{\text{cotxdx}}{\left\\{\sqrt\ (cos^{4}x+\ sin^{4}x)\right\\}} \right\rbrack
By taking $sin^4x$ as common from the denominator,
We get,
I=\int\left\lbrack \dfrac{\text{cotxdx}}{\left\\{\sqrt\ (sin^{4}x (\dfrac{{cos^{4}x}}{{sin^{4}x}}\ )+1)\right\\}}\right\rbrack
On taking sin⁴x outside the square root,
I = \int\left\lbrack \dfrac{\text{cotx dx}}{\left\\{ sin^{2}x {\left\\{\sqrt\ ((\dfrac{{cos^{4}x}}{sin^{4}x}\ )+1) \right\\}}\right\\}} \right\rbrack
We know that $\dfrac{{cosx}}{{sinx}} = cotx$
From this we can write,
$\dfrac{{cos}^{4}x}{{sin}^{4}x}\ = cot^{4}x$
I = \int\left\lbrack \dfrac{\text{cotx dx}}{\left\\{ sin^{2}x {\left\\{\sqrt\ {(cot^{4}+1) }\right\\}}\right\\}} \right\rbrack
Now, we also know that $\dfrac{1}{{sinx}} = cosecx$
On squaring both sides,
We get,
$\dfrac{1}{{sin}^{2}x} = cosec^{2}x$
By substituting this,
We get,
I = \int\left\lbrack \dfrac{{cotx\ cosec}^{2}{x\ dx}}{\left\\{\\{\sqrt\ {(cot^{4}+1) } \right\\}} \right\rbrack (1)
Now, we can take $cotx = t$
On differentiating both sides,
We get,
$- 2cotx\ cosec^{2}x\ dx\ = dt$
⇒ ${cotx cosec}^{2}x\ dx = - \dfrac{1}{2}{dt}$
By substituting this value in (1)
$I = \int\left( - \dfrac{1}{2}\dfrac{1}{\sqrt\ {(t^{2}+1)}}{dt} \right)$
On integrating both sides,
We get,
$I = - \dfrac{1}{2}|log\ \left( t +{\sqrt\ {(t^{2}+1)}}{dt} \ \right)|\ \ + c$
Now can substitute $cotx$ in the place of $t$,
We get,
I = - \dfrac{1}{2}\log\ \left| \cot^{2}x +{\sqrt\\{{((cot^{2}x}})^{2}+1)\right| + c
On simplifying,
We get,
I = \ \left(- \dfrac{1}{2} \right)\log\left| cotx\ + {\sqrt\\{(cot^{4}}x+1)\right| + c
Final answer :
\int\left\lbrack \dfrac{\text{cotxdx}}{\left\\{\sqrt\ (cos^{4}x+\ sin^{4}x) \ \right\\}} \right\rbrack =\ \left(- \dfrac{1}{2} \right)\log\left| cotx\ + {\sqrt\\{(cot^{4}}x+1)\right| + c
Option B). $\ \left(- \dfrac{1}{2} \right)\log\left| cot^{2}x\ + \ \ {\sqrt\ (cot^{4} + 1}) \right|$

Note:
The concept used in this question is integration method, that is integration by substitution . Since this is an indefinite integral we have to add an arbitrary constant ‘$c$’. $c$ is called the constant of integration. The variable $x$ in $dx$ is known as the variable of integration or integrator.