The correct option is B sin x log|tanx/2| x+c ∫ cosx log (tanx/2)dx = log(tanx/2) ∫ cosx dx ∫ 1/2 sec ^2 x/_2tan x/2 ∫ cosx dx =log ( ...

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Integration of Irrational Algebraic Fractions - 2

∫cos xlogtanx...

Question

$\int cos x log (tan \frac{x}{2}) d x =$

sin x log | tan x | - x + c

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- sin x log | tan \frac{x}{2} | + x + c

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- sin x log | tan \frac{x}{2} | - x + c

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sin x log | tan \frac{x}{2} | - x + c

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Solution

The correct option is B $sin x log | tan \frac{x}{2} | - x + c$
$\int c o s x l o g (t a n x / 2) d x$
$= l o g (t a n x / 2) \int c o s x d x$
$- \int 1 / 2 \frac{s e c^{2} x /_{2}}{t a n x / 2} \int c o s x d x$
$= l o g (t a n x / 2) s i n x - \int c o s e c x s i n x d x$
$= l o g (t a n x / 2) s i n x - x + c$
$= l o g (t a n x / 2) s i n x - x + c = \int c o s x l o g (t a n x / 2)^{d x}$
$= s i n x l o g | t a n \frac{x}{2} | - x + c$
Hence, option 'D' is correct.

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