Question

$\int \frac{1}{si{n}^{4}x+co{s}^{4}x}dx$

• A. $=\frac{1}{2\sqrt{2}}ta{n}^{-1}\left(\frac{ta{n}^{2}x-1}{\sqrt{2}tanx}\right)+C$
• B. $=\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{ta{n}^{2}x-1}{\sqrt{2}tanx}\right)+C$
• C. $=\frac{1}{\sqrt{2}}ta{n}^{-1}\left(d\frac{ta{n}^{2}x+1}{\sqrt{2}tanx}\right)+C$
• D. $=\frac{1}{\sqrt{3}}ta{n}^{-1}\left(\frac{ta{n}^{2}x-1}{\sqrt{2}tanx}\right)+C$

Answer 1

Answer: (B)

$=\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{ta{n}^{2}x-1}{\sqrt{2}tanx}\right)+C$

$I=\int \frac{1}{si{n}^{4}x+co{s}^{4}x}dx=\int \frac{1/co{s}^{4}x}{\frac{si{n}^{4}x+co{s}^{4}x}{co{s}^{4}x}}$
$=\int \frac{se{c}^{4}x}{ta{n}^{4}x+1}dx=\int \frac{se{c}^{2}xse{c}^{2}x}{ta{n}^{4}x+1}dx$
$=\int \left(\frac{1+ta{n}^{2}x}{1+ta{n}^4}x\right)se{c}^{2}xdx$
Substitute tan $x=t$ and $se{c}^{2}xdx=dt$, we get
$I=\int \frac{1+t^2}{1+t^4}dt=\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{t^2-1}{\sqrt{2}t}\right)+C$
$=\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{ta{n}^{2}x-1}{\sqrt{2}tanx}\right)+C$