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Question

The value of 1cosxcosαcosxdx, where 0<α<x<π, is equal to

A
2tan1(tan2x2+cosα1)+C
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B
2tan1(tan2x2(1+cosα)+cosα1)+C
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C
tan1tan2x2+cosα1252+C
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D
2tan1tan2x2(1+cosα)+cosα12+C
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Solution

The correct option is D 2tan1tan2x2(1+cosα)+cosα12+C
Consider the given integral.
I=1cosxcosαcosxdx
I=1cosxcosαcosxdx
I=   11tan2x21+tan2x2   cosα1tan2x21+tan2x2dx

Put t=tanx2
t=tanx2dt=sec2x2×12dxdx=2sec2x2dtdx=21+tan2x2dtdx=21+t2dt

Therefore,
I=11t21+t2cosα1t21+t221+t2dt
I=232t(t2+1)cosα+t2cosα1+t2dt
I=232t(t2+1)t2(cosα+1)+cosα1dt

Let, m=t2(1+cosα)+cosα1

dt=t2(1+cosα)+cosα1t(cosα+1)dm

Therefore,
I=2321m2+2dm
I=2322tan1m2+C
I=2tan1m2+C

Put the value of m, we get
I=2tan1t2(1+cosα)+cosα12+C

Put the value of t, we get
I=2tan1tan2x2(1+cosα)+cosα12+C
I=2tan1tan2x2(1+cosα)+cosα12+C

Hence, this is the value of the given integral.

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