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: Understanding the Concept
Combine the two integrals under one integral sign.

Step 2: Add
\[ I_n+I_{n+2}=\int_0^{\pi/4}\tan^nx\,(1+\tan^2x)\,dx=\int_0^{\pi/4}\tan^nx\sec^2x\,dx \]

Step 3: Substitute
Let \(t=\tan x\), \(dt=\sec^2x\,dx\). When \(x=0\), \(t=0\). When \(x=\pi/4\), \(t=1\).
\[ \int_0^1t^n\,dt=\frac{1}{n+1} \]

Step 4: Conclusion
\(I_{n+2}+I_n=\dfrac{1}{n+1}\), option (B). The other options have different denominators that do not come from \(\int t^n\,dt\).

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