Question

The value of $\int {\sec }^{4}x\cdot \tan xdx$ equals

• A. $\tan x+{\tan }^{3}x+c$
• B. $\frac{{\tan }^{2}x}{2}+\frac{{\tan }^{4}x}{4}+c$
• C. $\frac{{\tan }^{2}x}{2}+\frac{{\tan }^{3}x}{3}+c$
• D. $\tan x+\frac{{\tan }^{3}x}{3}+c$

Answer 1

Answer: (B)

$\frac{{\tan }^{2}x}{2}+\frac{{\tan }^{4}x}{4}+c$

$se{c}^{2}x=1+ta{n}^{2}x$

$\int (1+ta{n}^{2}x)tanxse{c}^{2}xdx$

$\int (1+t^2)tdt$ tanx=t dt=$se{c}^{2}x$

$\int (t+t^3)dt$

$\frac{t^2}{2}+\frac{t^4}{4}+c$

put t =tanx

$\frac{ta{n}^{2}x}{2}+\frac{ta{n}^{4}x}{4}+c$