Browse Questions

# Evaluate the following $\int\tan^2 x\sec^4 x\;dx$

$\begin{array}{1 1} (A)\;\large\frac{\tan ^2 x}{2}+\frac{\tan ^4 x}{4}+c \\ (B)\;\tan x+\frac{\tan ^2 x}{2}+c \\ (C)\;\large\frac{\tan ^3 x}{3}-\frac{\tan ^5 x}{5}+c \\ (D)\;\large\frac{\tan ^3 x}{3}+\frac{\tan ^5 x}{5}+c \end{array}$

Toolbox:
• If $f(x)$ is substituted by t, then $f'(x)dx=f'(t)dt.$ Hence $\int f(x)dx=\int t.dt$
• $f(x)=\tan x$ then $f(x)dx=sec^2xdx$
$\int\tan^2 x\sec^4 x\;dx=\int \tan^2 x.\sec^2 x.\sec ^2 x dx$
But $\sec^2 x=(1+\tan^2x)=\int \tan^2x(1+\tan^2x).sec^2xdx$
$=\int \tan^2x.\sec^2x\; dx+\int \tan^4 x.\sec^2xdx$
Put $\tan x=t$
Differentiting with respect to t
$\sec ^2 x dx=dt$
Now substituting in I we get,
$I=\int t^2.dt+\int t^4.dt$
on integrating we get,
$I=\large\frac{t^3}{3}+\frac{t^5}{5}+c$
Substituting for t we get
$I=\large\frac{\tan ^3 x}{3}+\frac{\tan ^5 x}{5}+c$