Question

$\int \frac{sin\left(2x\right)}{si{n}^2\left(x\right)+2co{s}^2\left(x\right)}dx=?$

• A. $–\log (1+{\sin }^{2}x)+c$
• B. $\log |1+{\cos }^{2}x|+c$
• C. $–\log |1+{\cos }^{2}x|+c$
• D. $\log (1+{\tan }^{2}x)+c$

Answer 1

The correct option is C

$–\log |1+{\cos }^{2}x|+c$

Explanation For The Correct Option:

Evaluating the integral:

$\int \frac{\text{sin}\left(2x\right)}{{\text{sin}}^2\left(x\right)+2{\text{cos}}^2\left(x\right)}dx$

$$\begin{gathered} \Rightarrow \int \frac{\text{sin}\left(2x\right)}{1-{\cos }^2\left(x\right)+2{\cos }^2\left(x\right)}dx[\because {\sin }^2\left(x\right)=1-{\cos }^2(x\left)\right] \\ \Rightarrow \int \frac{sin\left(2x\right)}{1+{\cos }^2\left(x\right)}dx \end{gathered}$$

Substituting $t=1+{\cos }^2\left(x\right)$ and also substituting $dx$ after differentiating $t$ w. r. t $x$

$$\begin{gathered} \Rightarrow dt=-2\sin \left(x\right)\cos \left(x\right)dx \\ \Rightarrow -2\sin \left(2x\right)dx=dt[\because \sin (x)=2\sin (x)\cos (x\left)\right] \end{gathered}$$

$$\begin{gathered} \Rightarrow -\int \frac{dt}{t}=-\log \left|t\right|+C[C\text{is integrating constants}] \\ =-\log |1+{\cos }^2(x\left)\right|+C[substitutingt=1+{\cos }^2(x\left)\right] \end{gathered}$$

Hence, option (C) is the correct answer.