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

The correct option is C π/4 1/2log 2∫_0^π/2 cos x/1+cos x+sin xdx =∫_0^π/2cos^2(x/2) sin^2(x/2)/2 cos^2(x/2)+2 sin(x/2)cos(x/2)dx =1/2∫_0^π/21 tan^2 (x/2)/1+ta ...

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Byju's Answer

Standard XII

Mathematics

First Fundamental Theorem of Calculus

∫0π/2cos x/1+...

Question

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

\frac{π}{4} + \frac{1}{2} l o g 2

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\frac{π}{4} + l o g 2

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\frac{π}{4} - \frac{1}{2} l o g 2

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\frac{π}{4} - l o g 2

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Solution

The correct option is C $\frac{π}{4} - \frac{1}{2} l o g 2$
$\int_{0}^{\frac{π}{2}} \frac{c o s x}{1 + c o s x + s i n x} d x$
$= \int_{0}^{\frac{π}{2}} \frac{c o s^{2} (\frac{x}{2}) - s i n^{2} (\frac{x}{2})}{2 c o s^{2} (\frac{x}{2}) + 2 s i n (\frac{x}{2}) c o s (\frac{x}{2})} d x$
$= \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{1 - t a n^{2} (\frac{x}{2})}{1 + t a n (\frac{x}{2})} d x = \frac{1}{2} \int_{0}^{\frac{π}{2}} [1 - t a n (\frac{x}{2})] d x$
$\frac{π}{4} + l o g \frac{1}{\sqrt{2}} = \frac{π}{4} - \frac{1}{2} l o g 2$

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Similar questions

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∫π20 cos x1+cos x+sin xdx=

The correct option is C π4−12log 2∫π20 cos x1+cos x+sin xdx =∫π20cos2(x2)−sin2(x2)2 cos2(x2)+2 sin(x2)cos(x2)dx =12∫π201−tan2(x2)1+tan(x2)dx=12∫π20[1−tan(x2)]dx π4+log 1√2=π4−12log 2

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