Question

How do you find the integral of \[\int {(1 + \cos x} {)^2}dx\]?

Answer 1

Hint:We can solve this using the algebraic identity \[{(a + b)^2} = {a^2} + {b^2} + 2ab\]. The term inside the integral sign is called the integrand. First we expand the integrand using the identity then we integrate. Where \[a = 1\] and \[b = \cos x\]. We know the integration of \[{x^n}\] with respect to ‘x’, that is \[\int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c\]. Where ‘c’ is the integration constant.

Complete step by step solution:
Given \[\int {(1 + \cos x} {)^2}dx\].
Now applying the identity \[{(a + b)^2} = {a^2} + {b^2} + 2ab\], where \[a = 1\] and \[b = \cos x\].
\[{\left( {1 + \cos x} \right)^2} = 1 + {\cos ^2}x + 2\cos x\]
Thus we have,
\[\int {(1 + \cos x} {)^2}dx = \int {(1 + {{\cos }^2}x + 2\cos x} )dx\]
\[ = \int {(1 + {{\cos }^2}x + 2\cos x} )dx\]
Expanding the integration for each term,
\[\int {(1 + \cos x} {)^2}dx = \int 1 .dx + \int {{{\cos }^2}x.} dx + \int {2\cos x.} dx\]
We know that \[{\cos ^2}x = \dfrac{{1 + \cos 2x}}{2}\], then
\[\int {{{(\cos x)}^2}.} dx = \int {\dfrac{{1 + \cos 2x}}{2}.dx} \]
\[ = \int {\dfrac{1}{2}} .dx + \int {\dfrac{{\cos 2x}}{2}.dx} \]
\[ = \dfrac{x}{2} + \dfrac{{\sin 2x}}{{2 \times 2}} + c\]
\[\int {{{(\cos x)}^2}.} dx = \dfrac{x}{2} + \dfrac{{\sin 2x}}{4} + c{\text{ }} - - - (1)\].
Also \[\int 1 .dx = x + c{\text{ }} - - - (2)\]
\[\int {2\cos x.} dx = 2\sin x + c{\text{ }} - - - (3)\]
Where ‘c’ is the integration constant.
Substituting (1) (2) and (3) in the above integral equation we have,
\[\int {(1 + \cos x} {)^2}dx = x + \dfrac{x}{2} + \dfrac{{\sin 2x}}{4} + 2\sin x + 3c\]
Since ‘3c’ is also a constant we denote this by “C”. Then we have
\[\int {(1 + \cos x} {)^2}dx = x + \dfrac{x}{2} + \dfrac{{\sin 2x}}{4} + 2\sin x + C\]
Where ‘C’ is the integration constant

Note: In the given above problem we have an indefinite integral, that is no upper and lower limit. Hence we add the integration constant ‘c’ after integrating. In a definite integral we will have an upper and lower limit, we don’t need to add integration constant in the case of definite integral. We have different integration rule:
The power rule: If we have a variable ‘x’ raised to a power ‘n’ then the integration is given by \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c} \].
The constant coefficient rule: if we have an indefinite integral of \[K.f(x)\], where f(x) is some function and ‘K’ represent a constant then the integration is equal to the indefinite integral of f(x) multiplied by ‘K’. That is \[\int {K.f(x)dx = c\int {f(x)dx} } \].
The sum rule: if we have to integrate functions that are the sum of several terms, then we need to integrate each term in the sum separately. That is
\[\int {\left( {f(x) + g(x)} \right)dx = \int {f(x)dx} } + \int {g(x)dx} \]
For the difference rule we have to integrate each term in the integrand separately.