- e x 2-sin 2 x -1-cos 2 x dx -A- e x cos 2 x - cB- 2 ex 2 x-cC- e x x - cD- e x tan x - c

The correct option is D e^xtan x+c I=∫ e^x[sec^2x+tan x ]dx=e^xtan x+c ...

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Byju's Answer

Standard XII

Mathematics

Composition of Trigonometric Functions and Inverse Trigonometric Functions

∫ e x 2+sin 2...

Question

$\int e^{x} (\frac{2 + s i n 2 x}{1 + c o s 2 x}) d x =$

$e^{x} c o t x + c$

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

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

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$e^{x} t a n x + c$

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Solution

The correct option is D
$e^{x} t a n x + c$

$I = \int e^{x} [s e c^{2} x + t a n x] d x = e^{x} t a n x + c$

Suggest Corrections

Similar questions

Q. 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)

\int e^{x} (1 - \cot x + \cot^{2} x) d x =

(a) e^x cot x + C
(b) −e^x cot x + C
(c) e^x cosec x + C
(d) −e^x cosec x + C

$\int e^{x} (\frac{1 - c o t x}{s i n x}) d x$

$\int \frac{x + 3}{{(x + 4)}^{2}} e^{x} d x =$

(a) $\frac{e^{x}}{x + 4} + C$

(b) $\frac{e^{x}}{x + 3} + C$

(c) $\frac{1}{{(x + 4)}^{2}} + C$

(d) $\frac{e^{x}}{{(x + 4)}^{2}} + C$

\int e^{x} (\frac{1 + sin x}{1 + cos x}) d x

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∫ex(2+sin 2x1+cos 2x)dx=

The correct option is D extan x+c I=∫ex[sec2x+tan x]dx=extan x+c

$\int e^{x} (\frac{2 + s i n 2 x}{1 + c o s 2 x}) d x =$

The correct option is D
$e^{x} t a n x + c$

$I = \int e^{x} [s e c^{2} x + t a n x] d x = e^{x} t a n x + c$