evaluate

IIT IS OBJECTIVE

SO PUT N=0 AND SOLVEANY ONE CAN DO THAT
IF 2 OPTIONS MATHES
PUT N=1 TO GET THE ANSWER

NOW THE PROBABILITY OF GETTING THE PERFECT OPTION INCREASES CONSIDERABLY[3]

IF FURTHER DOUBTS PUT N=2

take x=tanθ

∂x=sec²Î¸

i.e.,
∫tan^θ*sec^(2-n)θ∂θ

={tan^(n+1)θ/(n+1)}*sec^(-n)θ + ∫tan^(n+1)θ/n(n+1)}*sec^(-n-1)θ∂θ

=I +∫tan^(n+1)θ/n(n+1)}*sec^(-n-1)θ∂θ
=I+∫sin^(n+1)θ/(n*(n+1))

terefore we need to find ∫sin ^(n+1) θ∂θ

In=sin^nθδθ=∫sin^(n-2)θ*(1-cos²Î¸)δθ
=I(n-2) -∫sin^(n-2)θcos²Î¸Î´Î¸
Use by parts to see that this reduces to a reduction form

In=I(n-2) +In/(n-1) - cosθ*sin^(n-1)/(n-1)

Use this in the previous post

(I may be wrong in the calculations but this is the method)

In iMcq use what manipal has said