Question

$\int \frac{\sin 2x}{{\sin }^{4}x+{\cos }^{4}x}dx=?$

• A. $2{\tan }^{-1}({\tan }^{2}x)+c$
• B. ${\tan }^{-1}(x{\tan }^{2}x)+c$
• C. ${\tan }^{-1}({\tan }^{2}x)+c$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation For The Correct Option:

Evaluating the integral:

$\int \frac{\sin 2x}{{\sin }^{4}x+{\cos }^{4}x}dx$

$$\begin{gathered} \Rightarrow \int \frac{\sin 2x}{{\left({\sin }^{2}x\right)}^2+{\left({\cos }^{2}x\right)}^2} \\ \Rightarrow \int \frac{\sin 2x}{{\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-2{\sin }^{2}x{\cos }^{2}x}dx\left[\because {\left(a+b\right)}^2=a^2+b^2+2ab\Rightarrow a^2+b^2={\left(a+b\right)}^2-2ab\right] \\ \Rightarrow \int \frac{\sin 2x}{1-2{\sin }^{2}x{\cos }^{2}x}dx\left[\because {\sin }^{2}x+{\cos }^{2}x=1\right] \\ \Rightarrow \int \frac{\sin 2x}{1-\left({\displaystyle \frac{{\left(\sin 2x\right)}^2}{2}}\right)}dx\left[\because \sin 2x=2\mathrm{si}nx\cos x\right] \\ \Rightarrow 2\int \frac{\sin 2x}{2-{\left(\sin 2x\right)}^2}dx \\ \Rightarrow 2\int \frac{\sin 2x}{1+{\cos }^{2}2x}dx\left[\because {\sin }^2\theta =1-{\cos }^2\theta \right] \end{gathered}$$

Substituting $t=\cos \left(2x\right)$ then also putting value of $dx$ after differentiating $t$ w.r.t. $x$,

$dt=-2\sin 2xdx$

$$\begin{gathered} =-\int \frac{dt}{1+t^2} \\ =-{\tan }^{-1}t+C[C\text{is integrating constant}] \\ =-{\tan }^{-1}\left(\cos \left(2x\right)\right)+c\left[\text{substituting back}t=\cos \left(2x\right)\right] \end{gathered}$$

Therefore, $\int \frac{\sin 2x}{{\sin }^{4}x+{\cos }^{4}x}dx={\tan }^{-1}\left(\cos \left(2x\right)\right)+c$

Hence, option (D) is the correct answer.