Question

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

• 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)

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