If Iₙ=₀^{π / 4} ⁿ x d x, n N then Iₙ+2}+Iₙ is equal to

Correct Answer is : 1n+11/n+1n+11

Correct Answer is : 1n+11/n+1n+11

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MHTCET 20 Apr 2026 Shift 2 Paper

Question : 134 of 150

Marks: +1, -0

I_{n}=\int_{0}^{\pi / 4} \tan ^{n} x d x, n \in N

I_{n+2}+I_{n}

is equal to

Solution:

Concept:

Combine the integrals and simplify using the identity

1+\tan^2 x = \sec^2 x

Explanation:

We have

I_n = \int_{0}^{\pi/4} \tan^n x \, dx

So,

I_{n+2} + I_n = \int_{0}^{\pi/4} \tan^{n+2} x \, dx + \int_{0}^{\pi/4} \tan^n x \, dx

Combining the integrals gives

I_{n+2} + I_n = \int_{0}^{\pi/4} \left(\tan^{n+2}x + \tan^n x\right) dx.

Factor out

\tan^n x

I_{n+2} + I_n = \int_{0}^{\pi/4} \tan^n x \left(1+\tan^2 x\right) dx.

Using

1+\tan^2 x = \sec^2 x

, we get

I_{n+2} + I_n = \int_{0}^{\pi/4} \tan^n x \sec^2 x \, dx.

Let

t = \tan x

, so

dt = \sec^2 x \, dx

When

x=0

t=0

; when

x=\pi/4

t=1

Thus,

I_{n+2} + I_n = \int_{0}^{1} t^n \, dt

Evaluating the integral:

\int_{0}^{1} t^n dt = \left[\frac{t^{n+1}}{n+1}\right]_{0}^{1} = \frac{1}{n+1}.

Therefore,

I_{n+2} + I_n = \frac{1}{n+1}

Answer:

Option B:

\frac{1}{n+1}

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