Question

$\int \frac{\sin 2x}{1+{\cos }^{4}x}dx$ is equal to

• A. ${\cos }^{-1}\left({\cos }^{2}x\right)+c$
• B. ${\sin }^{-1}\left({\cos }^{2}x\right)+c$
• C. ${\cot }^{-1}\left({\cos }^{2}x\right)+c$
• D. $Noneofthese$

Answer 1

The correct option is A ${\cos }^{-1}\left({\cos }^{2}x\right)+c$

$\int \frac{sin2x}{1+co{s}^{4}x}dx=?$

Let $co{s}^{2}x=t$

diff.w.r.to x

$-2cosxsinxdx=dt$

$-sin2xdx=dt$ $(\because 2sinxcosx=sin2x)$

$\Rightarrow \int \frac{-dt}{1+t^2}$ $(\because \int \frac{-1}{1+x^2}dx=co{s}^{-1}(x)+c$

$\Rightarrow co{s}^{-1}\left(t\right)+c$

$\Rightarrow co{s}^{-1}\left(co{s}^{2}x\right)+c$