-x-sin x-1-cos xdx-x tan x-2 -p log - sec x-2 -c- p-

I=∫ ( x/2sec^2x/2 tanx/2 )dx I=x tanx/2 4 log |secx/2|+c ⇒ p nbsp;= 4 ...

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Byju's Answer

Standard XII

Mathematics

Integration by Partial Fractions

∫x-sin x/1+co...

Question

$\int \frac{x - s i n x}{1 + c o s x} d x = x t a n (\frac{x}{2}) + p l o g ∣ ∣ s e c (\frac{x}{2}) ∣ ∣ + c \Rightarrow p =$

-4

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-2

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Solution

The correct option is A
-4

$I = \int (\frac{x}{2} s e c^{2} \frac{x}{2} - t a n \frac{x}{2}) d x$
$I = x t a n \frac{x}{2} - 4 l o g ∣ ∣ s e c \frac{x}{2} ∣ ∣ + c$
$\Rightarrow p = - 4$

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∫x−sin x1+cos xdx=x tan(x2)+p log∣∣sec(x2)∣∣+c⇒p=

The correct option is A -4 I=∫(x2sec2x2−tanx2)dx I=x tanx2−4 log∣∣secx2∣∣+c ⇒p=−4

$\int \frac{x - s i n x}{1 + c o s x} d x = x t a n (\frac{x}{2}) + p l o g ∣ ∣ s e c (\frac{x}{2}) ∣ ∣ + c \Rightarrow p =$

The correct option is A
-4

$I = \int (\frac{x}{2} s e c^{2} \frac{x}{2} - t a n \frac{x}{2}) d x$
$I = x t a n \frac{x}{2} - 4 l o g ∣ ∣ s e c \frac{x}{2} ∣ ∣ + c$
$\Rightarrow p = - 4$