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Question: How do you integrate \[\dfrac{1}{{1 + \sin x}}\] ?...

How do you integrate 11+sin⁡x\dfrac{1}{{1 + \sin x}} ?

Explanation

Solution

In this problem of integral we will multiply numerator and denominator by 1−sin⁡x1 - \sin x .After that we will change the denominator by using the standard identity 1−sin⁡2x=cos⁡2x1 - {\sin ^2}x = {\cos ^2}x. Then we will split the terms in two separate terms and then we will find the integral of them separately.

Complete step by step answer:
Given that
∫11+sin⁡xdx\int {\dfrac{1}{{1 + \sin x}}dx}
Now we will multiply the numerator and denominator by 1−sin⁡x1 - \sin x.
=∫11+sin⁡x×1−sin⁡x1−sin⁡xdx= \int {\dfrac{1}{{1 + \sin x}} \times \dfrac{{1 - \sin x}}{{1 - \sin x}}dx}
In denominator we can see the form (a+b)(a−b)=a2−b2\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}
So rewriting the equation we get,
=∫1−sin⁡x12−sin⁡2xdx= \int {\dfrac{{1 - \sin x}}{{{1^2} - {{\sin }^2}x}}dx}
We know that 1−sin⁡2x=cos⁡2x1 - {\sin ^2}x = {\cos ^2}x
=∫1−sin⁡xcos⁡2xdx= \int {\dfrac{{1 - \sin x}}{{{{\cos }^2}x}}dx}
Separating the terms,
=∫1cos⁡2xdx+∫−sin⁡xcos⁡2xdx= \int {\dfrac{1}{{{{\cos }^2}x}}dx + \int {\dfrac{{ - \sin x}}{{{{\cos }^2}x}}dx} }
We know that 1cos⁡2x=sec⁡2x\dfrac{1}{{{{\cos }^2}x}} = {\sec ^2}x
=∫sec⁡2xdx+∫−sin⁡xcos⁡2xdx= \int {{{\sec }^2}xdx + \int {\dfrac{{ - \sin x}}{{{{\cos }^2}x}}dx} }
Also here sec⁡2xdx=tan⁡x{\sec ^2}xdx = \tan x
=tan⁡x+∫−sin⁡xcos⁡2xdx= \tan x + \int {\dfrac{{ - \sin x}}{{{{\cos }^2}x}}dx}
For finding the integral of the second term we will use the substitution.
Put u=cos⁡xu = \cos x
Taking derivative on both sides,
du=−sin⁡xdxdu = - \sin xdx
Now re-subtitute the values in the integral above,
=tan⁡x+∫1u2du= \tan x + \int {\dfrac{1}{{{u^2}}}} du
Here we know that ∫xndx=xn+1n+1\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}
So we will rewrite the integral above,
=tan⁡x+∫u−2du= \tan x + \int {{u^{ - 2}}} du
Using the formula above,
=tan⁡x+u−1−1= \tan x + \dfrac{{{u^{ - 1}}}}{{ - 1}}
Simplifying this term,
=tan⁡x−1u= \tan x - \dfrac{1}{u}
Re-substitute the value of u,
=tan⁡x−1cos⁡x= \tan x - \dfrac{1}{{\cos x}}
On solving the ratio,
=tan⁡x−sec⁡x+C= \tan x - \sec x + C
This is our final answer.

Note: Note that if students are facing the question in how to identify the method of solution it only comes with practice. For example in the above question the method used is substitution and multiplying the numerator and denominator by the same quantity in the denominator but with different signs. This is because we will get the standard identity sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1. After that we saw that the second integral has function and derivative both so we used substitution. Don’t forget to write the constant C after finding the integral.