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Question

sin2x+2tanxcos6x+6cos2x+4dx=

A
21+cos2xcos7x
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B
tan112(1+cos2xcos7x)
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C
112log(1+6cos4x+4cos6x)
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D
none
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Solution

The correct option is A 112log(1+6cos4x+4cos6x)
I=(cosx+1cosx)2sinxdxcos6x+6cos2x+4
Substitute cosx=t
sinxdx=dt
I=21+t2t7+6t3+4tdt
Divide above and below each term by t7
I=2(1t5+1t7)dt(1+6t4+4t6)
Substitute 1+6t4+4t6=z
24(1t5+1t7)dt=dz
I=112dzz=logz
Substituting the value of z
I=112log(1+6cos4x+4cos6x)

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