Question

If \(I_n = \int _0^{π/4}tan^nx dx,n\in N\) then \(I_{n+2}+I_n\) is equal to

• A. \(\frac{1}{n}\)
• B. \(\frac{1}{n+1}\)
• C. \(\frac{1}{n-1}\)
• D. \(\frac{1}{n-2}\)

Hint:

Add the two integrals and use sec squared as the derivative of tan.

Answer 1

The Correct Option is B

Step 1: Approach
Check with a value of $n$.

Step 2: Take $n=1$
$I_1=\int_0^{\pi/4}\tan x\,dx=\ln\sqrt2=\dfrac12\ln2$. $I_3=\int_0^{\pi/4}\tan^3x\,dx=\dfrac12-\dfrac12\ln2$ (since $\tan^3=\tan\sec^2-\tan$).

Step 3: Sum
$I_1+I_3=\dfrac12=\dfrac{1}{n+1}$ for $n=1$. For $n=1$ option (B) gives $\dfrac12$. Option (A) gives 1, option (C) is undefined and option (D) gives $-1$, so only (B) fits.

Step 4: Conclusion
Option (B).

Final Answer:
Adding the integrals gives the integral of t^n from 0 to 1, which is 1/(n+1), option (B). \[ \boxed{\frac{1}{n+1}} \]