Question:

Integrate the function: \(\frac {1}{1-tan \ x}\)

Updated On: Oct 19, 2023
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Solution and Explanation

\(Let \ I = ∫\frac {1}{1-tan \ x } dx\)

\(I= ∫\frac {1}{1-\frac {sin\  x}{cos \ x }}dx\)

\(I= ∫\frac {cos\  x}{cos \ x - sin\  x} dx\)

\(I= \frac 12 ∫\frac {2cos\  x}{cos\  x - sin\  x} dx\)

\(I= \frac 12 ∫\frac {(cos \ x-sin\  x)+(cos\  x+sin \ x)}{(cos \ x-sin\  x) }dx\)

\(I= \frac 12 ∫1dx+\frac 12 ∫\frac {cos \ x+sin\  x}{cos\  x-sin\  x} dx\)

\(I= \frac x2+\frac 12 ∫\frac {cos \ x+sin\  x}{cos\  x-sin\  x} dx\)

\(Put  \ cos\  x - sin\  x = t ⇒ (-sin\  x - cos \ x) dx = dt\)

\(∴ I = \frac x2+\frac 12 ∫\frac {-(dt)}{t}\)

\(I= \frac x2-\frac 12 log\ |t|+C\)

\(I= \frac x2-\frac 12 log\ |cos\ x-sin\ x|+C\)

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Concepts Used:

Methods of Integration

Given below is the list of the different methods of integration that are useful in simplifying integration problems:

Integration by Parts:

 If f(x) and g(x) are two functions and their product is to be integrated, then the formula to integrate f(x).g(x) using by parts method is:

∫f(x).g(x) dx = f(x) ∫g(x) dx − ∫(f′(x) [ ∫g(x) dx)]dx + C

Here f(x) is the first function and g(x) is the second function.

Method of Integration Using Partial Fractions:

The formula to integrate rational functions of the form f(x)/g(x) is:

∫[f(x)/g(x)]dx = ∫[p(x)/q(x)]dx + ∫[r(x)/s(x)]dx

where

f(x)/g(x) = p(x)/q(x) + r(x)/s(x) and

g(x) = q(x).s(x)

Integration by Substitution Method

Hence the formula for integration using the substitution method becomes:

∫g(f(x)) dx = ∫g(u)/h(u) du

Integration by Decomposition

Reverse Chain Rule

This method of integration is used when the integration is of the form ∫g'(f(x)) f'(x) dx. In this case, the integral is given by,

∫g'(f(x)) f'(x) dx = g(f(x)) + C

Integration Using Trigonometric Identities