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

Hint: 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.