Question

If $\int \frac{dx}{1+\sin x} = tan (\frac{x}{2}+a)+b$ , find the values of a and b.

Answer 1

$\sf \underline{Answer}$

Given:-

$\sf \int \frac{dx}{1+\sin x} = tan (\frac{x}{2}+a)+b$ – (i)

But,

$\sf \int \frac{dx}{1+\sin x}$

$\sf =\int \frac{dx}{1+\cos (\frac{π}{2}- x)}$

$\sf = \int \frac{dx}{1+\cos( x - \frac{π}{2})}$

$\sf = \int \frac{dx}{2\cos²( \frac{x}{2} - \frac{π}{4})}$

$\sf = \frac{1}{2} \sec ²( \frac{x}{2} - \frac{π}{4}) dx$

$\sf = \frac{1}{2} \frac{\tan ( \frac{x}{2} - \frac{π}{4})}{ \frac{1}{2} } + C$

$\sf \tan (\frac{x}{2} - \frac{π}{4}) + C – (ii)$

on comparing (i) and (ii),

$\sf \: a = \frac{ - \pi}{4} \: and \: b = C, any \: real \: number$

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