The correct option is C π ∫x1+sin xdx =∫xcos ^2x/1 sin xdx =∫ ( xcos ^2x x sin xcos ^2x )dx =x tan x +ln|cos x| xsin x tan x x cos x+ ...

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Property 2

0π∫x dx 1 + ...

Question

$π \int 0 \frac{x d x}{1 + sin x}$ =

\frac{π}{6}

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π

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\frac{π}{3}

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Cannot be valued

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Solution

The correct option is C $π$
$\int \frac{x}{1 + sin x} d x$

$= \int \frac{x}{\frac{{cos}^{2} x}{1 - sin x}} d x$

$= \int (\frac{x}{{cos}^{2} x} - \frac{x sin x}{{cos}^{2} x}) d x$

$= x tan x + ln | cos x | - x sin x tan x - x cos x + ln | tan x + sec x | + C$

$\int_{0}^{π} \frac{x}{1 + sin x} d x = {[x tan x + ln | cos x | - x sin x tan x - x cos x + ln | tan x + sec x |]}_{0}^{π}$

$\int_{0}^{π} \frac{x}{1 + sin x} d x = π$

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Similar questions

The maximum value of $s i n (x + \frac{π}{6}) + c o s (x + \frac{π}{6})$ in the

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