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Question

Solve: 2sin2 xcos x6cos2 x4sin xdθ.

A
2log(sin2xsinx+5)+7tan1(x2)+C
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B
2log(sin2xsin4x+5)7tan1(x2)+C
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C
log(sin2xsin4x+5)7tan1(x2)+C
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D
2log(sin2x+sinx+5)+7tan1(x2)+C
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Solution

The correct option is B 2log(sin2xsin4x+5)7tan1(x2)+C
2sin2θcosθ6cos2θ4sinθdθ
2sin2θcosθ4sinθcos2θ+6dθ
=2sin2θcosθ4sinθ+cos2θ6dθ
Substituting
sin2θ=2sinθcosθcos2θ=1sin2θ
=cosθ(4sinθ1sin2θ4sinθ+5)dθ
Substituting u=sinθdθ=1cosθdu
=[4u1u24u+5du]
4u1 as 2(2uu)+7
=(2(2u4)u24u+5+7u24u+5)du
=4u2u24u+5du+71u24u+5du
v=u24u+5du=12u4dv
=412vdv
=121vdv
=lnv2
=ln(u24u+5)2
=1(u2)2+1du
v=u2du=dv
=1v2+1dv
=tan1(v)
v=u2
=tan1(u2)
=4ln(u24u+5)2+7tan1(u2)
=2ln(u24u+5)7tan1(u2)
=2ln(sin2θ4sinθ+5)+7tan1(sinθ2)
2sin2θcosθ6cos2θ4sinθdθ=3ln(sin2θ4sinθ+5)+7tan1(sinθ2)

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