Question

integration 1/1+cot x dx
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Answer 1

HAVE A LOOK AT THE ATTACHMENT...HOPE IT HELPS✨

✯SHUKRAAN✯

Answer 2

Answer:

$\sf I =\dfrac{x}{2}-\dfrac{1}{2}\:log|sin\:x+cos\:x|+C$

Step-by-step explanation:

Given:

$\sf \dfrac{1}{1+cot\:x}$

To Find:

$\sf \int\limits {\dfrac{1}{1+cot\:x} } \, dx$

Solution:

$\sf Let\:I=\int\limits {\dfrac{1}{1+cot\:x} } \, dx$

We know,

cot x = cos x/sin x

$\sf I= \int\limits{\dfrac{1}{1+\frac{cos\:x}{sin\:x} } } \, dx$

$\sf = \int\limits{\dfrac{1}{\frac{sin\:x+cos\:x}{sin\:x} } } \, dx$

$\sf =\int\limits {\dfrac{sin\:x}{sin\:x+cos\:x} } \, dx$

Now multiplying and dividing by 2 we get,

$\sf =\int\limits \dfrac{2}{2} \times {\dfrac{sin\:x}{sin\:x+cos\:x} } \, dx$

Taking 1/2 outside the integral,

$\sf =\dfrac{1}{2} \int\limits {\dfrac{2\:sin\:x}{sin\:x+cos\:x} } \, dx$

Now add and subtract cos x on the numerator,

$\sf =\dfrac{1}{2} \int\limits {\dfrac{2\:sin\:x+cos\:x-cos\:x}{sin\:x+cos\:x} } \, dx$

$\sf =\dfrac{1}{2} \int\limits {\dfrac{\:sin\:x+cos\:x+sin\:x-cos\:x}{sin\:x+cos\:x} } \, dx$

Separating the terms,

$\sf =\dfrac{1}{2}\int\limits {\dfrac{sin\:x+cos\:x}{sin\:x+cos\:x} } \, dx +\dfrac{1}{2} \int\limits {\dfrac{sin\:x-cos\:x}{sin\:x+cos\:x} } \, dx$

$\sf = \dfrac{1}{2}\int\limits {1 } \, dx +\dfrac{1}{2} \int\limits {\dfrac{sin\:x-cos\:x}{sin\:x+cos\:x} } \, dx$

We know,

$\sf \int\limits {1} \, dx =x+C$

Hence,

$\sf =\dfrac{1}{2}\times x +\dfrac{1}{2} \int\limits {\dfrac{sin\:x-cos\:x}{sin\:x+cos\:x} } \, dx$

Let us assume,

t = sin x + cos x

Differentiating on both sides,

dt = cos x - sin x dx

-dt = sin x - cos x

Therefore,

$\sf I =\dfrac{x}{2} +\dfrac{1}{2} \int {\dfrac{-dt}{t} }$

$\sf I =\dfrac{x}{2} -\dfrac{1}{2} \int {\dfrac{dt}{t} }$

We know,

$\sf \int\limits {\dfrac{dt}{t} } = log|t| +C$

Hence,

$\sf I =\dfrac{x}{2}-\dfrac{1}{2}\:log|t|+C$

Substitute the value of t,

$\sf I =\dfrac{x}{2}-\dfrac{1}{2}\:log|sin\:x+cos\:x|+C$