integration of 1 sinx cosx

we can have three integration of 1 sinx cosx ( integration of 1 /sinx+ cosx , integration of 1 /sinx- cosx and integration of 1 /sinx cosx). Here are formula of these integrals

(I) integration of 1 /sinx+ cosx

$\int \frac {1}{sinx + cos x} \; dx = \frac {1}{\sqrt{2}} \log \left| \frac {\left(\tan\left(\frac{x}{2}\right) – 1 + \sqrt{2}\right)}{\left(\tan\left(\frac{x}{2}\right) -1- \sqrt{2} \right)}\right| + C$

or
$\int \frac {1}{sinx + cos x} \; dx = \frac {1}{\sqrt{2}} \ln |\tan ( \frac {x}{2} + \frac {\pi}{8}) | + C$

(II) integration of 1 /sinx- cosx

$\int \frac {1}{sinx – cos x} \; dx = \frac {1}{\sqrt 2} ln |\frac {\tan\left(\frac{x}{2}\right) +1- \sqrt 2}{\tan\left(\frac{x}{2}\right) + 2+ \sqrt 2}| + C$

or
$\int \frac {1}{sinx – cos x} \; dx = \frac {1}{\sqrt{2}} \ln |\tan ( \frac {x}{2} – \frac {\pi}{8}) | + C$

(III) integration of 1 /sinx cosx

$\int \frac {1}{sin x cos x} \; dx = \frac{1}{2} \ln |\frac {(\cos(2x) – 1)}{ (\cos(2x) + 1)}| + C$

or

$\int \frac {1}{sin x cos x} \; dx= \ln |tan x| + C$

Proof of integration of 1/sinx+cosx

The integral can be tricky due to the combination of sine and cosine functions in the denominator. A common approach is to use a tangent half-angle substitution, which simplifies the integration of trigonometric functions that involve both sine and cosine.

Method 1

The tangent half-angle substitution is $ t = \tan\left(\frac{x}{2}\right) $. This substitution leads to the identities:

$$
\sin x = \frac{2t}{1 + t^2}, \quad \cos x = \frac{1 – t^2}{1 + t^2}, \quad dx = \frac{2}{1 + t^2} dt
$$

Substituting these into the integral, we get:

$$
\int \frac{1}{\sin x + \cos x} dx = \int \frac{1}{\frac{2t}{1 + t^2} + \frac{1 – t^2}{1 + t^2}} \cdot \frac{2}{1 + t^2} dt
$$

$$
=\int \frac{2}{2t + 1 – t^2} \; dt
$$

$$
=\int \frac{2}{2 – (t^2 + 1 -2t)} \; dt
$$

$$
=\int \frac{2}{(\sqrt 2)^2 – (t-1)^2} \; dt
$$

Now we know that
$\int \frac {1}{a^2 – x^2} dx = \frac {1}{2a} ln |\frac {a+x}{a-x}| + C$
Proof
$\frac {1}{a^2 – x^2} =\frac {1}{2a}[ \frac {1}{a-x} + \frac {1}{a+x}]$
So
$\int \frac {1}{a^2 – x^2} dx $
$=\frac {1}{2a}[ \int \frac {1}{a-x} dx + \int \frac {1}{x+a}]$
$= \frac {1}{2a}[-ln |a-x| + ln |a+x| + C$
$=\frac {1}{2a} ln |\frac {a+x}{a-x}| + C$

Therefore

$$
=\frac {1}{\sqrt{2}} \left( \log(t – 1 + \sqrt{2}) – \log(t – \sqrt{2} – 1) \right) + C
$$

where $ t = \tan\left(\frac{x}{2}\right) $ and $ C $ is the constant of integration.

To express this in terms of $ x $, we substitute back $ t = \tan\left(\frac{x}{2}\right) $, resulting in the final solution:

$$
=\frac {1}{\sqrt{2}} \log |\frac {\left(\tan\left(\frac{x}{2}\right) – 1 + \sqrt{2}\right)}{\left(\tan\left(\frac{x}{2}\right) – \sqrt{2} – 1\right)}| + C
$$

Method 2

$\int \frac {1}{sinx + cos x} \; dx = \int \frac {1}{ \sqrt 2 sin (x + \frac {\pi}{4})} \; dx$
$=\frac {1}{\sqrt{2}} \int \csc (x + \frac {\pi}{4}) \; dx$

Now
\[
\int \csc(x) \, dx = \ln | \tan \frac {x}{2} | + C
\]

Therefore, original integral becomes

$=\frac {1}{\sqrt{2}} \ln |\tan ( \frac {x}{2} + \frac {\pi}{8}) | + C$

Proof of integration of 1/sinx- cosx

The integral can be tricky due to the combination of sine and cosine functions in the denominator. A common approach is to use a tangent half-angle substitution, which simplifies the integration of trigonometric functions that involve both sine and cosine.

Method 1

The tangent half-angle substitution is $ t = \tan\left(\frac{x}{2}\right) $. This substitution leads to the identities:

$$
\sin x = \frac{2t}{1 + t^2}, \quad \cos x = \frac{1 – t^2}{1 + t^2}, \quad dx = \frac{2}{1 + t^2} dt
$$

Substituting these into the integral, we get:

$$
\int \frac{1}{\sin x – \cos x} dx = \int \frac{1}{\frac{2t}{1 + t^2} – \frac{1 – t^2}{1 + t^2}} \cdot \frac{2}{1 + t^2} dt
$$

$$
=\int \frac {2}{2t – 1 + t^2} \; dt
$$

$$
=\int \frac {2} {(t^2 + 1 -2t) – 2} \; dt
$$

$$
=\int \frac{2} {(t+1)^2 – (\sqrt 2)^2} \; dt
$$

$\int \frac {1}{x^2 – a^2} dx = \frac {1}{2a} ln |\frac {x-a}{x+a}| + C$
Proof
$\frac {1}{x^2 – a^2} =\frac {1}{2a}[ \frac {1}{x-a} – \frac {1}{x+a}]$
So
$\int \frac {1}{x^2 – a^2} dx $
$=\frac {1}{2a}[ \int \frac {1}{x-a} dx – \int \frac {1}{x+a}]$
$= \frac {1}{2a}[ln |x-a| – ln |x+a| + C$
$=\frac {1}{2a} ln |\frac {x-a}{x+a}| + C$

Therefore

$$
=\frac {1}{\sqrt 2} ln |\frac {t+1- \sqrt 2}{t + 2+ \sqrt 2}| + C
$$

To express this in terms of $ x $, we substitute back $ t = \tan\left(\frac{x}{2}\right) $, resulting in the final solution:

$$
=\frac {1}{\sqrt 2} ln |\frac {\tan\left(\frac{x}{2}\right) +1- \sqrt 2}{\tan\left(\frac{x}{2}\right) + 2+ \sqrt 2}| + C
$$

Method 2

$\int \frac {1}{sinx – cos x} \; dx = \int \frac {1}{ \sqrt 2 sin (x – \frac {\pi}{4})} \; dx$
$=\frac {1}{\sqrt{2}} \int \csc (x – \frac {\pi}{4}) \; dx$

Now
\[
\int \csc(x) \, dx = \ln | \tan \frac {x}{2} | + C
\]

Therefore, original integral becomes

$=\frac {1}{\sqrt{2}} \ln |\tan ( \frac {x}{2} – \frac {\pi}{8}) | + C$

Proof of integration of 1/sinx cosx

Method 1

$\int \frac {1}{sin x cos x} \; dx = \int \frac {2}{ sin 2x} \; dx $
$= 2 \int \csc 2x \; dx$

Now
\[
\int \csc(x) \, dx = \ln | \tan \frac {x}{2} | + C
\]

Therefore, original integral becomes

$= \ln |tan x| + C $

Method 2

We also know that

\[
\int \csc(x) \, dx = \frac {1}{2} \ln | \frac {\cos x -1}{\cos x + 1} | + C
\]

Therefore, original integral becomes

$= \ln |\frac {(\cos(2x) – 1)}{ (\cos(2x) + 1)}| + C $

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