Question

$\int \frac{sinxcosx}{si{n}^{4}x+co{s}^{4}x}dx=$

• A. $ta{n}^{-1}\left(sinx\right)+c$
• B. $\frac{1}{2}ta{n}^{-1}\left(sin2x\right)+c$
• C. $\frac{1}{2}ta{n}^{-1}\left(ta{n}^{2}x\right)+c$
• D. $\frac{1}{2}ta{n}^{-1}\left(co{t}^{2}x\right)+c$

Answer 1

The correct option is C $\frac{1}{2}ta{n}^{-1}\left(ta{n}^{2}x\right)+c$

$\int \frac{sinxcosx}{si{n}^{4}x+co{s}^{4}x}dx=\int \frac{tanx.se{c}^{2}x}{ta{n}^{4}x+1}dx$
Put $ta{n}^{2}x=t\Rightarrow 2tanx.se{c}^{2}xdx=dt=\frac{1}{2}\int \frac{1}{t^2+1}dt=\frac{1}{2}ta{n}^{-1}t+c=\frac{1}{2}ta{n}^{-1}\left(ta{n}^{2}x\right)+c$