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

The correct option is C π/4 log √(2)I = ∫_0^π/2sin x/2/sin x/2 + cosx/2 dx = 2 ∫_0^π/4sin x/sin x + cos x dx Let sin x = A (sin x + cos x) + B(cos x sin x) ...

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Byju's Answer

Standard XII

Mathematics

Area between x=g(y) and y Axis

∫0π/2sin x/1+...

Question

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

\frac{π}{4}

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\frac{π}{4} + l o g \sqrt{2}

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\frac{π}{4} - l o g \sqrt{2}

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\frac{π}{4} - l o g 2

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Solution

The correct option is C $\frac{π}{4} - l o g \sqrt{2}$
$I = \int_{0}^{\frac{π}{2}} \frac{s i n \frac{x}{2}}{s i n \frac{x}{2} + c o s \frac{x}{2}} d x = 2 \int_{0}^{\frac{π}{4}} \frac{s i n x}{s i n x + c o s x} d x$
Let $s i n x = A (s i n x + c o s x) + B (c o s x - s i n x)$
A - B = 1 and A + B = 0,
$A = \frac{1}{2}, B = \frac{- 1}{2} I = 2 \times \frac{1}{2} \times \frac{π}{4} - 2 \times \frac{1}{2} [l o g (s i n x + c o s x)]_{0}^{\frac{π}{4}} = \frac{π}{4} - l o g \sqrt{2}$

Suggest Corrections

∫π20sinx1+cosx+sinxdx=

The correct option is C π4−log√2I=∫π20sinx2sinx2+cosx2dx=2∫π40sinxsinx+cosxdx Let sinx=A(sinx+cosx)+B(cosx−sinx) A - B = 1 and A + B = 0, A=12,B=−12I=2×12×π4−2×12[log(sinx+cosx)]π40=π4−log√2

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