Question

The $\int {\sqrt {1 + \sin \left( {\dfrac{x}{4}} \right)} } \,dx$ equals:
A. $8\left[ {\sin \left( {\dfrac{x}{8}} \right) + \cos \left( {\dfrac{x}{8}} \right)} \right] + c$
B. $8\left[ {\sin \left( {\dfrac{x}{8}} \right) - \cos \left( {\dfrac{x}{8}} \right)} \right] + c$
C. $8\left[ {\sin \left( {\dfrac{x}{8}} \right) - \sin \left( {\dfrac{x}{8}} \right)} \right] + c$
D. $\dfrac{1}{8}\left[ {\sin \left( {\dfrac{x}{8}} \right) - \cos \left( {\dfrac{x}{8}} \right)} \right] + c$

Answer 1

Hint: In the given question, we have to evaluate the integral involving the square root of a trigonometric expression. So, first we will eliminate the square root of the function by converting the function inside into a whole square. We will use algebraic identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$. Trigonometric identities such as ${\sin ^2}x + {\cos ^2}x = 1$.

Complete step by step answer:
The given question requires us to integrate a function in x: $\sqrt {1 + \sin \left( {\dfrac{x}{4}} \right)} $ with respect to x. So, we have,
$\int {\sqrt {1 + \sin \left( {\dfrac{x}{4}} \right)} } \,dx$
Now, we know the trigonometric identity ${\sin ^2}x + {\cos ^2}x = 1$. So, we can write ${\sin ^2}\left( {\dfrac{x}{8}} \right) + {\cos ^2}\left( {\dfrac{x}{8}} \right)$ as one. So, we get,
$ \Rightarrow \int {\sqrt {{{\sin }^2}\left( {\dfrac{x}{8}} \right) + {{\cos }^2}\left( {\dfrac{x}{8}} \right) + \sin \left( {\dfrac{x}{4}} \right)} } \,dx$

Now, we use the double angle formula of sine as $\sin 2x = 2\sin x\cos x$. Hence, we get,
$ \Rightarrow \int {\sqrt {{{\sin }^2}\left( {\dfrac{x}{8}} \right) + {{\cos }^2}\left( {\dfrac{x}{8}} \right) + 2\sin \left( {\dfrac{x}{8}} \right)\cos \left( {\dfrac{x}{8}} \right)} } \,dx$
Now, we use the algebraic identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ to condense the expression.
$ \Rightarrow \int {\sqrt {{{\left[ {\sin \left( {\dfrac{x}{8}} \right) + \cos \left( {\dfrac{x}{8}} \right)} \right]}^2}} } \,dx$
So, we get,
$ \Rightarrow \int {\sin \left( {\dfrac{x}{8}} \right) + \cos \left( {\dfrac{x}{8}} \right)} \,dx$

Now, we can use the chain rule of integration \[\int {\sin \left( {ax} \right)\,} dx = - \dfrac{{\cos \left( {ax} \right)}}{a} + c\] for linear expressions. We also know the integral of $\sin \theta $ as $ - \cos \theta $ and the integral of $\cos \theta $ as $\sin \theta $. Hence, we get,
$ \Rightarrow \left[ { - \dfrac{{\cos \left( {\dfrac{x}{8}} \right)}}{{\left( {\dfrac{1}{8}} \right)}} + \dfrac{{\sin \left( {\dfrac{x}{8}} \right)}}{{\left( {\dfrac{1}{8}} \right)}}} \right]\,$
Simplifying the expression, we get,
\[ \Rightarrow 8\sin \left( {\dfrac{x}{8}} \right) - 8\cos \left( {\dfrac{x}{8}} \right) + c\,\], where c is any arbitrary constant.
Factoring out the common factors, we get,
\[ \therefore 8\left[ {\sin \left( {\dfrac{x}{8}} \right) - \cos \left( {\dfrac{x}{8}} \right)} \right] + c\,\]

So, the integral $\int {\sqrt {1 + \sin \left( {\dfrac{x}{4}} \right)} } \,dx$ equals \[8\left[ {\sin \left( {\dfrac{x}{8}} \right) - \cos \left( {\dfrac{x}{8}} \right)} \right] + c\,\] where c is any arbitrary constant.

Note: The indefinite integrals of certain functions may have more than one answer in different forms. However, all these forms are correct and interchangeable into one another. Indefinite integral gives us the family of curves as we don’t know the exact value of the arbitrary constant, c. We must know the double angle formula of sine and the chain rule of integration for linear expressions to solve the given integral.