\(∫1/(1 + sinx)?\)
Solution and Explanation
∫1/(1 + sinx)dx
∫1/(1 + sin x)·(1 - sin x)/(1 - sin x)dx
= ∫(1 − sin x)/(1 − sin2x)dx
From trigonometric identities, we know that sin2x + cos2x = 1.
From this, cos2x = 1 – sin2x
Substituting this in the above integral,
= ∫(1 − sin x)/cos2xdx
= ∫(1/cos2x) - (sin x)/(cos x)·(1/cosx)dx
= ∫ (sec2x – tan x sec x)dx
= tan x − sec x + C (∵ ∫ sec2x dx = tan x and ∫ tan x sec x dx = sec x)
Thus, ∫1 / (1 + sin x) dx = tan x − sec x + C, where C is the integration constant.
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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