Question

If $\int {\sec }^{4}xdx=\frac{{\sec }^{m}x\tan x}{n}+k\tan x+C$
$(C$ is integration constant$)$, then the value of $m+2n+3k$ is equal to

Answer 1

$I=\int {\sec }^{4}xdx=\int {\underset{I}{\sec }}^{2}x\cdot {\underset{II}{\sec }}^{2}xdx$
Applying integration by parts
$\Rightarrow I={\sec }^{2}x\cdot \tan x-\int \tan x\cdot 2\sec x\cdot \sec x\tan xdx={\sec }^{2}x\cdot \tan x-2\int {\sec }^{2}x({\sec }^{2}x-1)dxI={\sec }^{2}x\cdot \tan x-2I+2\int {\sec }^{2}xdx\Rightarrow 3I={\sec }^{2}x\cdot \tan x+2\tan x+C\Rightarrow I=\frac{{\sec }^{2}x\cdot \tan x}{3}+\frac{2}{3}\tan x+C\therefore m=2,n=3,k=\frac{2}{3}$
$\Rightarrow m+2n+3k=2+6+2=10$