The correct option is A cos 2x4+c ∫dx/tanx + cotx = ∫dx/sinx/cosx+cosx/sinx #160; #160; = ∫cosx sinx/sin^2x+cos^2xdx = 1/2∫2sinx.cos ...

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Logarithmic Differentiation

∫ dx tan x + ...

Question

$\int \frac{d x}{tan x + cot x} =$

- \frac{cos 2 x}{4} + c

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\frac{sin 2 x}{4} + c

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\frac{- sin 2 x}{4} + c

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\frac{cos 2 x}{4} + c

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Solution

The correct option is A $- \frac{cos 2 x}{4} + c$
$\int \frac{d x}{t a n x + c o t x}$
$= \int \frac{d x}{\frac{s i n x}{c o s x} + \frac{c o s x}{s i n x}} = \int \frac{c o s x - s i n x}{s i n^{2} x + c o s^{2} x} d x$
$= \frac{1}{2} \int \frac{2 s i n x . c o s x}{1} d x$ $(∵ s i n^{2} x + c o s^{2} x = 1)$
$= \frac{1}{2} \int s i n (2 x) d x$
$= \frac{1}{2} \frac{- c o s (2 x)}{2} + c$
$= \frac{- c o s (2 x)}{4} + c$

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Similar questions

Q. Find

B

. If

\int \frac{s i n^{2} x - c o s^{2} x}{s i n^{2} x c o s^{2} x} d x = tan x + B + c

Q. Solve:

\int \frac{d x}{cos x \sqrt{1 + cos 2 x + sin 2 x}}; (0 < x < \frac{π}{4})

Q. If

tan x = - \frac{3}{4},

\frac{3}{2} π < x < 2 π

then find the
i)

sin 2 x

ii)

cos 2 x

iii)

tan 2 x

Q. Prove that

\frac{1 + s i n 2 x}{c o s 2 x} = t a n (\frac{π}{4} + 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
$(a)tan x+cot x+C (b)tan x+cosec x+C (c)-tan x+cot x+C (d)tan x+sec x+C$

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∫dxtanx+cotx=

The correct option is A −cos2x4+c∫dxtanx+cotx=∫dxsinxcosx+cosxsinx=∫cosx−sinxsin2x+cos2xdx=12∫2sinx.cosx1dx (∵sin2x+cos2x=1)=12∫sin(2x)dx=12−cos(2x)2+c=−cos(2x)4+c

$\int \frac{d x}{tan x + cot x} =$