Question

Please solve this integral ASAP
and please no fake answers
​

Answer 1

Question

$\sf \displaystyle\int \dfrac{dx}{ {sin}^{4}x + {cos}^{4} x } dx$

Solution with explaination:

$\sf\hookrightarrow \displaystyle\int \dfrac{dx}{ {sin}^{4}x + {cos}^{4} x } dx$

Now.,dividing both numerator and denominator by cos⁴x.

$\sf\hookrightarrow \displaystyle\int\dfrac{ \dfrac{1}{ {cos}^{4}x } }{ \dfrac{ {sin}^{4}x }{ {cos}^{4} x} + 1} dx$

$\sf \because\boxed{ \dfrac{1}{cosx} = secx}$

$\sf\hookrightarrow \displaystyle\int \dfrac{ {sec}^{4} x}{ {tan}^{4} x + 1} dx = \displaystyle \int\dfrac{ {sec}^{2}x \times {sec}^{2} x }{ {tan}^{4}x + 1 } dx$

$\sf\hookrightarrow\displaystyle \int \dfrac{ {sec}^{2} x(1 + {tan}^{2}x) }{1 + {tan}^{4}x } dx$

Let tanx =t

sec²xdx=dt

$\sf\hookrightarrow\displaystyle \int \dfrac{1 + {t}^{2} }{1 + {t}^{4} } dt$

Now ,divide both numerator and denominator by t².

$\sf\hookrightarrow\displaystyle \int \frac{1 + \dfrac{1}{ {t}^{2} } }{ {t}^{2} + \dfrac{1}{ {t}^{2} } } dt$

$\sf\hookrightarrow \displaystyle\int \dfrac{1 + \dfrac{1}{ {t}^{2} } }{ {t}^{2} + \dfrac{1}{ {t}^{2} } - 2 + 2} dt$

$\sf\hookrightarrow \displaystyle\int \dfrac{1 + \dfrac{1}{ {t}^{2} } }{ {(t - \frac{1}{t} )}^{2} + 2 } dt$

$let \: t - \dfrac{1}{t} = u$

$\sf\bigg(1 + \dfrac{1}{ {t}^{2} } \bigg)dt = du$

$\hookrightarrow \displaystyle\int \dfrac{du}{ {u}^{2} + 2 }$

$\sf\boxed{\bigg[\displaystyle \int \dfrac{dx}{ {a}^{2} + {x}^{2} } = \dfrac{1}{a} {tan}^{ - 1}\bigg (\dfrac{x}{a} \bigg) + C\bigg]}$

$\sf\hookrightarrow \displaystyle\int \frac{1}{ \sqrt{2} } {tan}^{ - 1} \bigg( \dfrac{u}{ \sqrt{2} } \bigg) + C$

$\sf\hookrightarrow\displaystyle \int \dfrac{1}{ \sqrt{2} } {tan}^{ - 1} \bigg( \frac{t - \dfrac{1}{t} }{ \sqrt{2} } \bigg) + C$

$\sf\hookrightarrow \displaystyle\int \dfrac{1}{ \sqrt{2} } {tan}^{ - 1} \bigg( \dfrac{ {t}^{2} - 1}{ \sqrt{2} t}\bigg ) + C$

$\sf= \dfrac{1}{ \sqrt{2} } {tan}^{ - 1}\bigg ( \dfrac{ {tan}^{2}x - 1 }{ \sqrt{2}tanx }\bigg ) + C$

Answer 2

$\sf \huge \underline{ \red{AnSwEr }}:-$