Question

$\int\left(\sec ^{4} x+\tan ^{4} x\right) d x=$

• A. $\frac{2}{3} \tan ^{3} x-\frac{2}{3} \tan x +x+ c$
• B. $\frac{1}{3} \sec ^{2} x \tan x+\frac{5}{3} \tan x+\frac{\tan ^{3} x}{3}+x +c$
• C. $\frac{2}{3} \tan ^{3} x+ x+ c$
• D. $\frac{1}{3} \sec ^{2} x \tan x-\frac{5}{3} \tan x+\frac{\tan ^{3} x}{3}+x +c$

Answer 1

Answer: (C)

$I=\int\left(\sec ^{4} x+\tan ^{4} x\right) d x$
$=\int\left[\sec ^{4} x+\left(\sec ^{2} x-1\right)^{2}\right] d x$
$=\int\left(2 \sec ^{4} x-2 \sec ^{2} x+1\right) d x$
$=\int\left[2\left(1+\tan ^{2} x\right) \sec ^{2} x-2 \sec ^{2} x+1\right] d x$
$=\int\left(2 \tan ^{2} x \sec ^{2} x+1\right) d x=\frac{2}{3} \tan ^{3} x +x +c$