Question

Let${\mathrm{I}}_{\mathrm{n}}=\int {\tan }^{\mathrm{n}}\mathrm{xdx}(\mathrm{n}>1)$. If ${\mathrm{I}}_4+{\mathrm{I}}_6={\mathrm{atan}}^5\mathrm{x}+{\mathrm{bx}}^5+\mathrm{C}$, where $\mathrm{C}$ is a constant of integration, then the ordered pair $(\mathrm{a},\mathrm{b})$ is equal to

• A. $(1/5,0)$
• B. $(1/5,-1)$
• C. $(-1/5,0)$
• D. $(-1/5,1)$

Answer 1

The correct option is A

$(1/5,0)$

Determine the ordered pair $(\mathrm{a},\mathrm{b})$

Given, ${\mathrm{I}}_{\mathrm{n}}=\int {\tan }^{\mathrm{n}}\mathrm{xdx}(\mathrm{n}>1)$.

$$\begin{gathered} \therefore {\mathrm{I}}_{\mathrm{n}}+{\mathrm{I}}_{\mathrm{n}+2}=\int {\tan }^{\mathrm{n}}\left(\mathrm{x}\right)d\mathrm{x}+\int \tan {\left(\mathrm{x}\right)}^{\mathrm{n}+2}d\mathrm{x} \\ =\int {\tan }^{\mathrm{n}}\left(\mathrm{x}\right)(1+\tan {\left(\mathrm{x}\right)}^2)d\mathrm{x} \\ =\int {\tan }^{\mathrm{n}}\left(\mathrm{x}\right){\sec }^2\left(\mathrm{x}\right)d\mathrm{x} \\ =\frac{{\tan }^{\mathrm{n}+1}\left(\mathrm{x}\right)}{\mathrm{n}+1}+\mathrm{C} \end{gathered}$$

Put $\mathrm{n}=4$we get,

$$\begin{gathered} {\mathrm{I}}_{\mathrm{n}}+{\mathrm{I}}_{\mathrm{n}+2}=\frac{{\tan }^{\mathrm{n}+1}\left(\mathrm{x}\right)}{\mathrm{n}+1}+\mathrm{C} \\ \Rightarrow {\mathrm{I}}_4+{\mathrm{I}}_6=\frac{{\tan }^5\left(\mathrm{x}\right)}{5}+\mathrm{C} \\ \Rightarrow {\mathrm{I}}_4+{\mathrm{I}}_6=\frac{1}{5}{\tan }^5\left(\mathrm{x}\right)+0\times {\mathrm{x}}^5+\mathrm{C} \end{gathered}$$

Therefore, comparing with the result we can see that $\mathrm{a}=\frac{1}{5},\mathrm{b}=0$. Ordered pair is $\left(\frac{1}{5},0\right)$

Hence, option A is the correct answer.