The correct option is C x tan x+C ∫cos 2x 1cos 2x+1dx ⇒∫1 2 sin^2x 12cos^2x 1+1dx ⇒∫ tan^2x dx ⇒∫ 1 ^2dx ⇒ x tan x+c . ...

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The Problem of Areas

∫ cos 2x-1cos...

Question

$\int \frac{c o s 2 x - 1}{c o s 2 x + 1} d x$

t a n x - C

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x + t a n x + C

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x - t a n x + C

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- x - c o t x + C

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Solution

The correct option is C $x - t a n x + C$
$\int \frac{cos 2 x - 1}{cos 2 x + 1} d x$
$\Rightarrow \int \frac{1 - 2 {sin}^{2} x - 1}{2 {cos}^{2} x - 1 + 1} d x$
$\Rightarrow \int - {tan}^{2} x d x$
$\Rightarrow \int 1 - {sec}^{2} d x$
$\Rightarrow x - tan x + c$ .

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Similar questions

\int \frac{\cos 2 x - 1}{\cos 2 x + 1} d x =

(a) tan x − x + C
(b) x + tan x + C
(c) x − tan x + C
(d) − x − cot x + C

is equal to

A. tan x + cot x + C

B. tan x + cosec x + C

C. − tan x + cot x + C

D. tan x + sec x + C

Q. If

\int e^{sin x} [\frac{x {cos}^{3} x - sin x}{{cos}^{2} x}] d x = e^{sin x} f (x) + c

where c is constant of integration, then

f (x) =

Choose the correct answer. $\int \frac{s i n^{2} x - c o s^{2} x}{s i n^{2} x c o s^{2} x} d x$ is equal to
$(a)tan x+cot x+C (b)tan x+cosec x+C (c)-tan x+cot x+C (d)tan x+sec x+C$

Q. The integral of

\int \frac{1 - sin x}{{cos}^{2} x} d x

(a) tan x + sec x + c (b) tan x - sec x + c

(a) sec x - tan x + c (b) tan x - ln sec x + c

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∫cos2x−1cos2x+1dx

The correct option is C x−tanx+C∫cos2x−1cos2x+1dx⇒∫1−2sin2x−12cos2x−1+1dx⇒∫−tan2xdx⇒∫1−sec2dx⇒x−tanx+c.

$\int \frac{c o s 2 x - 1}{c o s 2 x + 1} d x$

The correct option is C $x - t a n x + C$
$\int \frac{cos 2 x - 1}{cos 2 x + 1} d x$
$\Rightarrow \int \frac{1 - 2 {sin}^{2} x - 1}{2 {cos}^{2} x - 1 + 1} d x$
$\Rightarrow \int - {tan}^{2} x d x$
$\Rightarrow \int 1 - {sec}^{2} d x$
$\Rightarrow x - tan x + c$ .