Question

A: If $I_n=\int {\tan }^{n}xdx$, then $5(I_4+I_6)={\tan }^{5}x$.
R: lf $I_n=\int {\tan }^{n}xdx$, $I_n=\frac{{\tan }^{n-1}x}{n}+I_{n-2}$, where $n\in N$.

• A. Both A and R are true and R is the correct explanation of A.
• B. Both A and R are true but R is not correct explanation of A.
• C. A is true R is false.
• D. A is false but R is true.

Answer 1

Answer: (C)

A is true R is false.

$I_n=\int {\tan }^{n}xdx$
$I_n=\int {\tan }^{n-2}xta{n}^{2}xdx$
$=\int {\tan }^{n-2}x[{\sec }^{2}x-1]\cdot dx$
$I_n=\int {\tan }^{n-2}xd\left(\tan x\right)-\int {\tan }^{n-2}xdx$
$I_n+I_{n-2}=\int {\tan }^{n-2}xd\left(\tan x\right)=\frac{{\tan }^{n-1}x}{n-1}+c$, where $c$ is the constant of integration
$5(I_6+I_4)={\tan }^{5}x+c$
Therefore, A is true but R is false.
Hence, option C is correct answer.