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: Use the tangent reduction in a handy form.
From $\int\tan^n x\,dx=\dfrac{\tan^{n-1}x}{n-1}-\int\tan^{n-2}x\,dx$, we get $I_k+I_{k-2}=\dfrac{\tan^{k-1}x}{k-1}$ for $k\ge2$.
Step 2: Regroup the given combination.
Write $I_0+I_1+2I_2+2I_3+2I_4+I_5+I_6$ as $(I_0+I_2)+(I_1+I_3)+(I_2+I_4)+(I_3+I_5)+(I_4+I_6)$, since splitting the coefficient $2$ pairs each middle term twice.
Step 3: Replace each pair by the reduction result.
Using $I_k+I_{k-2}=\dfrac{\tan^{k-1}x}{k-1}$ on $(I_2+I_0),(I_3+I_1),(I_4+I_2),(I_5+I_3),(I_6+I_4)$ gives $\tan x,\ \dfrac{\tan^2x}{2},\ \dfrac{\tan^3x}{3},\ \dfrac{\tan^4x}{4},\ \dfrac{\tan^5x}{5}$.
Step 4: Add them up.
The sum is $\tan x+\dfrac{\tan^2x}{2}+\dfrac{\tan^3x}{3}+\dfrac{\tan^4x}{4}+\dfrac{\tan^5x}{5}$.
Step 5: Match to the given series.
This is exactly $\displaystyle\sum_{k=1}^{n}\dfrac{\tan^k x}{k}$ with the last term $\dfrac{\tan^5x}{5}$.
Step 6: Read off $n$.
The top index is $5$, so $n=5$.
\[ \boxed{5} \]