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MHTCET 26 Apr 2025 Shift 1 Paper

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Question : 142 of 150

Marks: +1, -0

\int_{\frac{1}{2}}^{2} \frac{1}{x} \operatorname{cosec}^{101}\left(x-\frac{1}{x}\right) d x=

0
1
$\frac{1}{4}$
$\frac{101}{2}$

Solution:

Concept:

Use the substitution

x = \frac{1}{t}

and the odd property of

\operatorname{cosec}

to show that the integral equals its own negative.

Explanation:

Let

I = \int_{\frac{1}{2}}^{2} \frac{1}{x} \operatorname{cosec}^{101}\left(x-\frac{1}{x}\right) dx

.

Put

x = \frac{1}{t}

, so

dx = -\frac{1}{t^2}dt

.

When

x = \frac{1}{2}

,

t = 2

; when

x = 2

,

t = \frac{1}{2}

.

Then

I = \int_{2}^{\frac{1}{2}} -\frac{1}{t}\operatorname{cosec}^{101}\left(\frac{1}{t}-t\right)dt.

Since

\operatorname{cosec}\left(\frac{1}{t}-t\right) = -\operatorname{cosec}\left(t-\frac{1}{t}\right)

and

101

is odd,

\operatorname{cosec}^{101}\left(\frac{1}{t}-t\right) = -\operatorname{cosec}^{101}\left(t-\frac{1}{t}\right).

Therefore

I = \int_{2}^{\frac{1}{2}} \frac{1}{t}\operatorname{cosec}^{101}\left(t-\frac{1}{t}\right)dt = -\int_{\frac{1}{2}}^{2}\frac{1}{x}\operatorname{cosec}^{101}\left(x-\frac{1}{x}\right)dx = -I.

So

I = -I \Rightarrow 2I = 0 \Rightarrow I = 0

.

Answer:

0

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