Question

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

Answer 1

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.