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Question

List IList II(A)If(x2+cos2x1+x2)cosec2x dx=mcot1x +n cosec xsecx,then(P)m=1(B)Ifx+x2+2 dx=m3(x+x2+2)3/2(Q)n=1 nx+x2+2,then(C)Ifxtan1x(1+x2)3/2dx=mx1+x2(R)n=2 +ntan1x1+x2,then(D)Ifsin2xsin4x+cos4xdx=ncot1(tan2x) +msin2x,then(S)m=1(T)m=0(U)n=1

Which of the following is the only INCORRECT combination?

A
(A)(P),(Q)
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B
(B)(P),(R)
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C
(C)(S),(R)
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D
(D)(T),(Q)
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Solution

The correct option is C (C)(S),(R)
(A)
I=(x2+cos2x1+x2)cosec 2x dx
=(x2+1sin2x1+x2)cosec 2x dx
=(cosec 2x11+x2)dx
=cotx+cot1x+k
=cosec xsecx+cot1x+k
m=1,n=1
(A)(P),(Q)

(B)
I=x+x2+2 dx
Put x2+2+x=t
x2+2x=2t
2x=t2t2dx=(1+2t2)dt
I=12(t+2t3/2)dt
I=12t3/23/2+112t+k
=13(x+x2+2)3/22x+x2+2+k
m=1,n=2
(B)(P),(R)

(C)
I=xtan1x(1+x2)3/2dx=xtan1x(1+x2)1+x2dx
Put x=tanθ, where θ(π2,π2)
dx=sec2θ dθ
dx1+x2=dθ
So, I=θtanθ1+tan2θdθ
=θtanθsecθdθ
=θsinθ dθ
=θcosθ+cosθ=θcosθ+sinθ+k
Now, x=tanθ
cosθ=11+x2 and sinθ=x1+x2
I=xtan1x1+x2
Hence, m=1,n=1
(C)(P),(Q)

(D)
sin2xsin4x+cos4xdx
Dividing numerator and denominator by cos4x,
I=2tanxsec2xtan4x+1dx
Put tan2x=t2tanxsec2x dx=dt
I=dtt2+1
=tan1(t)+k1
=cot1(t)+k, where k=π2+k1
=cot1(tan2x)+k
m=0,n=1
(D)(T),(Q)

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