-0-2x-sin x-1-cos x d x-

The correct option is C π/2∫_0^π/2x+sin x/1+cos xdx=∫_0^π/2x+sin x/2cos^2 x/2dx =1/2∫_0^π/2x sec^2 x/2dx+∫_0^π/2tan x/2dx. = | x tan x/2 |^π/2_0=π/2tan π/4=π/2 ...

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Byju's Answer

Standard XII

Mathematics

First Fundamental Theorem of Calculus

∫0π/2x+sin x/...

Question

$\int_{0}^{\frac{π}{2}} \frac{x + s i n x}{1 + c o s x} d x =$

-log 2

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log 2

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\frac{π}{2}

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0.0

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Solution

The correct option is C $\frac{π}{2}$
$\int_{0}^{\frac{π}{2}} \frac{x + s i n x}{1 + c o s x} d x = \int_{0}^{\frac{π}{2}} \frac{x + s i n x}{2 c o s^{2} \frac{x}{2}} d x$
$= \frac{1}{2} \int_{0}^{\frac{π}{2}} x s e c^{2} \frac{x}{2} d x + \int_{0}^{\frac{π}{2}} t a n \frac{x}{2} d x .$
$= {∣ ∣ x t a n \frac{x}{2} ∣ ∣}_{0}^{\frac{π}{2}} = \frac{π}{2} t a n \frac{π}{4} = \frac{π}{2}$

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∫π20x+sin x1+cos xdx=

The correct option is C π2∫π20x+sin x1+cos xdx=∫π20x+sin x2cos2 x2dx =12∫π20x sec2x2dx+∫π20tan x2dx. =∣∣x tan x2∣∣π20=π2tanπ4=π2

$\int_{0}^{\frac{π}{2}} \frac{x + s i n x}{1 + c o s x} d x =$