Question

If $\int {\tan }^{4}xdx=a{\tan }^{3}x+b\tan x+\varphi \left(x\right)$, then

• A. $a=\frac{1}{3}$
• B. $b=1$
• C. $\varphi \left(x\right)=x+C$
• D. $b=-1$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $a=\frac{1}{3}$
C $\varphi \left(x\right)=x+C$
D $b=-1$

$\int {\tan }^{4}xdx=a{\tan }^{3}x+b\tan x+\varphi \left(x\right)$ ....(1)
Now $\int {\tan }^{4}xdx$
$=\int {\tan }^{2}x({\sec }^{2}x-1)dx$
$=\int {\tan }^{2}x{\sec }^{2}xdx-\int {\tan }^{2}xdx$
Put $\tan x=t$
${\sec }^{2}xdx=dt$
$=\int t^{2}dt-\int ({\sec }^{2}x-1)dx$
$=\frac{t^3}{3}-\tan x+x+C$
$=\frac{{\tan }^{3}x}{3}-\tan x+x+C$
So, on comparing with (1), we get
$a=\frac{1}{3},b=-1$
$\varphi \left(x\right)=x+C$