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Question: If \(\int_{}^{}\frac{dx}{1 + \sin x} = \tan\left( \frac{x}{2} + a \right) + b,\) then...

If ∫dx1+sin⁡x=tan⁡(x2+a)+b,\int_{}^{}\frac{dx}{1 + \sin x} = \tan\left( \frac{x}{2} + a \right) + b, then

A

a=π4,b=3a = \frac{\pi}{4},b = 3

B

a=−π4,b=3a = \frac{- \pi}{4},b = 3

C

a=π4,b=arbitrary constanta = \frac{\pi}{4},b = \text{arbitrary constant}

D

a=−π4,b=arbitrary constanta = \frac{- \pi}{4},b = \text{arbitrary constant}

Answer

a=−π4,b=arbitrary constanta = \frac{- \pi}{4},b = \text{arbitrary constant}

Explanation

Solution

If a=ba = b, then ∫dxa+bsin⁡x=−1acot⁡(π4+x2)+c\int_{}^{}\frac{dx}{a + b\sin x} = - \frac{1}{a}\cot\left( \frac{\pi}{4} + \frac{x}{2} \right) + c

∴\therefore ∫dx1+sin⁡x=−cot⁡(π4+x2)\int_{}^{}{\frac{dx}{1 + \sin x} = - \cot\left( \frac{\pi}{4} + \frac{x}{2} \right)} = −tan⁡(π2−π4−x2)+c- \tan\left( \frac{\pi}{2} - \frac{\pi}{4} - \frac{x}{2} \right) + c =

−tan⁡(π4−x2)+c- \tan\left( \frac{\pi}{4} - \frac{x}{2} \right) + c = tan⁡(x2−π4)+c\tan\left( \frac{x}{2} - \frac{\pi}{4} \right) + c

Hence a=−π4a = \frac{- \pi}{4} and b=b = arbitrary constant.