Question

solve the question step by step​

Answer 1

Solution:

Given Integral:

$\displaystyle\rm\longrightarrow I=\int\dfrac{\sin(2x)}{sin^{4}(x)+cos^{4}(x)}\:dx$

We know that:

$\rm\longrightarrow \sin(2x)=2\sin(x)\cos(x)$

$\rm\longrightarrow\sin(x)=\dfrac{\sin(x)}{\cos(x)}\times\cos(x)=\tan(x)\cos(x)=\dfrac{\tan(x)}{\sec(x)}$

$\rm\longrightarrow\cos(x)=\dfrac{1}{\sec(x)}$

Using this result, we get:

$\displaystyle\rm\longrightarrow I=\int\dfrac{2\sin(x)\cos(x)}{\dfrac{tan^{4}(x)}{\sec^{4}(x)}+\dfrac{1}{\sec^{4}(x)}}\:dx$

$\displaystyle\rm\longrightarrow I=\int\dfrac{2\sin(x)\cos(x)}{\dfrac{tan^{4}(x)+1}{\sec^{4}(x)}}\:dx$

$\displaystyle\rm\longrightarrow I=\int\dfrac{sec^{4}(x)\cdot2\sin(x)\cos(x)}{tan^{4}(x)+1}\:dx$

$\displaystyle\rm\longrightarrow I=\int\dfrac{sec^{3}(x)\cdot2\sin(x)}{tan^{4}(x)+1}\:dx$

$\displaystyle\rm\longrightarrow I=\int\dfrac{sec^{2}(x)\cdot2\dfrac{\sin(x)}{\cos(x)}}{tan^{4}(x)+1}\:dx$

$\displaystyle\rm\longrightarrow I=\int\sec^{2}(x)\cdot\dfrac{2\tan(x)}{tan^{4}(x)+1}\:dx$

Now, let us assume that:

$\rm\longrightarrow u=\tan(x)$

$\rm\longrightarrow \dfrac{du}{dx}=\sec^{2}(x)$

$\rm\longrightarrow dx=\dfrac{du}{\sec^{2}(x)}$

So, the integral changes to:

$\displaystyle\rm\longrightarrow I=\int\sec^{2}(x)\cdot\dfrac{2u}{u^{4}+1}\cdot\dfrac{du}{\sec^{2}(x)}$

$\displaystyle\rm\longrightarrow I=\int\cdot\dfrac{2u\:du}{u^{4}+1}$

Now, let us assume that:

$\rm\longrightarrow t=u^{2}$

$\rm\longrightarrow dt=2u\:du$

So, the integral changes to:

$\displaystyle\rm\longrightarrow I=\int\cdot\dfrac{dt}{t^{2}+1}$

$\displaystyle\rm\longrightarrow I=tan^{-1}(t)+C$

Substituting back the value of t, we get:

$\displaystyle\rm\longrightarrow I=tan^{-1}(u^{2})+C$

Substituting back the value of u, we get:

$\displaystyle\rm\longrightarrow I=tan^{-1}(tan^{2}x)+C$

Therefore:

$\displaystyle\rm\longrightarrow\int\dfrac{\sin(2x)}{sin^{4}(x)+cos^{4}(x)}\:dx=tan^{-1}(tan^{2}x)+C$

Which is our required answer.

Learn More:

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}$