JEE Main 25 Feb 2021 Shift 2 Solved Paper

In=

∫

‌ π 4

‌

π

2

cotnxdx=

∫

‌ π 4

‌

π

2

cotn−2x(cot2x)dx
In=

∫

‌ π 4

‌

π

2

cotn−2xcosec2xdx−

∫

‌ π 4

‌

π

2

cotn−2xdx
In+In−2=

∫

‌ π 4

‌

π

2

cotn−2x⋅cosec2xdx
Now, let cot‌x=t, then cosec2xdx=−dt, limit will be
In+In−2‌‌=

0

∫

1

−tn−2dt
‌=‌

−(t)n−1

n−1

]10=−{‌

0

n−1

−‌

(1)n−1

n−1

}
In+In−2=‌

1

n−1

Now, put n=4
⇒‌‌I2+I4=‌

1

3

, then ‌

1

I2+I4

=3 . . . (i)
Put n=5
⇒‌‌I5+I3=‌

1

4

, then ‌

1

I3+I5

=4. . . (ii)
Put n=6
⇒‌‌I6+I4=‌

1

5

, then ‌

1

I4+I6

=5. . . (iii)
Here, from Eqs. (i), (ii) and (iii), we conclude
‌

1

I2+I4

,‌

1

I3+I5

and ‌

1

I4+I6

are in AP with common difference 1.