Question

If
\[ I_n=\int \tan^n x\,dx \] and
\[ I_0+I_1+2I_2+2I_3+2I_4+I_5+I_6=\sum_{k=1}^{n}\frac{\tan^k x}{k}, \] then \(n=\)

• A. \(6\)
• B. \(5\)
• C. \(4\)
• D. \(3\)

Hint:

For integrals of powers of \(\tan x\), use \(I_n=\frac{\tan^{n-1}x}{n-1}-I_{n-2}\).

Answer 1

The Correct Option is B

Step 1: Write required integrals.
Using reduction formula,
\[ I_n=\frac{\tan^{n-1}x}{n-1}-I_{n-2} \]
We get,
\[ I_0=x \]
\[ I_1=\log(\sec x) \]
\[ I_2=\tan x-x \]
\[ I_3=\frac{\tan^2x}{2}-\log(\sec x) \]
\[ I_4=\frac{\tan^3x}{3}-\tan x+x \]
\[ I_5=\frac{\tan^4x}{4}-\frac{\tan^2x}{2}+\log(\sec x) \]
\[ I_6=\frac{\tan^5x}{5}-\frac{\tan^3x}{3}+\tan x-x \]

Step 2: Substitute in the given expression.
\[ I_0+I_1+2I_2+2I_3+2I_4+I_5+I_6 \]
On simplifying, the \(x\) terms and logarithmic terms cancel.
The remaining expression becomes
\[ \tan x+\frac{\tan^2x}{2}+\frac{\tan^3x}{3}+\frac{\tan^4x}{4}+\frac{\tan^5x}{5} \]

Step 3: Compare with the given summation.
\[ \sum_{k=1}^{n}\frac{\tan^k x}{k} = \tan x+\frac{\tan^2x}{2}+\frac{\tan^3x}{3}+\cdots+\frac{\tan^n x}{n} \]
Comparing, the last term is
\[ \frac{\tan^5x}{5} \]
Therefore,
\[ n=5 \]

Step 4: Final conclusion.
Hence,
\[ \boxed{5} \]