Question

$\frac{si{n}^{3}x}{(co{s}^{4}x+3co{s}^{2}x+1)ta{n}^{-1}(secx+cosx)}dx=$

• A. $ta{n}^{-1}(secx+cosx)+c$
• B. $log|ta{n}^{-1}(secx+cosx)|+c$
• C. $\frac{1}{(secx+co{s}^{2}x{)}^2}+c$
• D. $log|secx+cosx|+c$

Answer 1

Answer: (B)

$log|ta{n}^{-1}(secx+cosx)|+c$

Put $ta{n}^{-1}(secx+cosx)=f\left(x\right)f^1\left(x\right)=\frac{si{n}^{3}x}{co{s}^{4}x+3co{s}^{2}x+1}\therefore \int \frac{f^1\left(x\right)}{f\left(x\right)}=log|f\left(x\right)|+c$