Question

If $\int {\sec }^{4}x{\text{cosec}}^{2}xdx=\frac{1}{3}{\tan }^{3}x+\frac{A}{701}\tan x-\cot x+C$, then $A$ is equal to

Answer 1

Let $I=\int {\sec }^{4}x{\csc }^{2}xdx=\int {({\tan }^{2}x+1)}^2({\cot }^{2}x+2)dx$

Substituting $t=\tan x\Rightarrow dt={\sec }^{2}xdx$

Here $dx=\frac{dt}{{\sec }^{2}x}=\frac{dt}{1+{\tan }^{2}x}=\frac{dt}{1+t^2}$
Therefore, $I=\int (\frac{1}{t^2}+1)(t^2+1)dt=\int (t^2+\frac{1}{t^2}+2)dt$

$=\frac{t^3}{3}-\frac{1}{t}+2t+c$

$=\frac{{\tan }^{3}x}{3}-\cot x+2\tan x+c$

Thus, $A=1402$