Let f ( x ) = 2 s i n 2 x − 1 c o s x + c o s x ( 2 s i n x + 1 ) 1 + s i n x then ∫ e x ( f ( x ) + f ′ ( x ) ) d x equals (where c is the constant of integeration) Options: A . &nbsp e x t a n x + c B . &nbsp e x c o t x + c C . &nbsp e x c o s e c 2 x + c D . &nbsp None of these

Answer: Option A : A c o s x ( 1 + 2 s i n x ) 1 + s i n x − c o s 2 x − s i n 2 x c o s x = c o s 2 x ( 1 + 2 s i n x ) − ( 1 + s i n x ) ( c o s 2 x − s i n 2 x ) c o s x ( 1 + s i n x ) = s i n x c o s 2 x + s i n 2 x ( 1 + s i n x ) c o s x ( 1 + s i n x ) = s i n x ( 1 − s i n x ) + s i n 2 x c o s x = t a n x

Let f(x)=2sin2x−1cosx+cosx(2sinx+1)1+sinx then ∫ex(f(x)+f′

Discussion Forum : Indefinite Integration

Question -

Let $f (x) = \frac{2 s i n^{2} x - 1}{c o s x} + \frac{c o s x (2 s i n x + 1)}{1 + s i n x}$ then $\int e^{x} (f (x) + f^{'} (x)) d x$ equals
(where c is the constant of integeration)

Options:

A .

e^{x} t a n x + c

B .

e^{x} c o t x + c

C .

e^{x} c o s e c^{2} x + c

D . None of these

Answer: Option A
:
A

\frac{c o s x (1 + 2 s i n x)}{1 + s i n x} - \frac{c o s^{2} x - s i n^{2} x}{c o s x} = \frac{c o s^{2} x (1 + 2 s i n x) - (1 + s i n x) (c o s^{2} x - s i n^{2} x)}{c o s x (1 + s i n x)}

= \frac{s i n x c o s^{2} x + s i n^{2} x (1 + s i n x)}{c o s x (1 + s i n x)} = \frac{s i n x (1 - s i n x) + s i n^{2} x}{c o s x} = t a n x

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Discussion Forum : Indefinite Integration

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