Question

# If x satisfies the equation x2∫10dtt2+2tcosα+1−x∫3−3t3sin2tt2+1dt−2=0 , (0<α<π), then the value of x is?

A
2(sinαα)
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B
2(sinαα)
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C
4(sinαα)
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D
None of these
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Solution

## The correct options are B 2√(sinαα) C −2√(sinαα)Let,I1=∫10dtt2+2tcosα+1I2=∫3−3t2sin2tt2+1dt Now,I1=∫10dtt2+2tcosα+1=∫10dtt2+2tcosα+cos2α+1−cos2α=∫10dt(t+cosα)2+sin2α=1sinα[tan−1t+cosαsinα]10[Using ∫1x2+a2=1atan−1xa]=1sinα[tan−11+cosαsinα]−1sinα[tan−1cosαsinα]=1sinα⎡⎢ ⎢⎣tan−12cos2α22sinα2⋅cosα2⎤⎥ ⎥⎦−1sinα[tan−1cosαsinα]=1sinα[tan−1cotα2]−1sinα[tan−1cotα]Given that, 0<α<π∴cotα2=tan(π2−α2)andcotα=tan(π2−α)∴I1=1sinα[tan−1tan(π2−α2)]−1sinα[tan−1tan(π2−α)]=1sinα[(π2−α2)−(π2−α)]=1sinα[α2]=α2sinαNow,I2=∫3−3t2sin2tt2+1dt=0[∵t2sin2tt2+1 is an odd function] Thus, the given equation reduces to x2⋅α2sinα−2=0⇒x2⋅α2sinα=2⇒x2=4sinαα⇒x=±2√sinααHence, options (A) and (B) are correct.

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