Question

Evaluate \[\int {\cos \sqrt x dx} \]
A. \[\left[ {\sqrt x \sin \sqrt x + \cos \sqrt x } \right]\]
B. \[2\left[ {\sin \sqrt x - \cos \sqrt x } \right]\]
C. \[2\left[ {\sqrt x \sin \sqrt x + \sqrt x \cos \sqrt x } \right]\]
D. \[2\left[ {\sqrt x \sin \sqrt x + \cos \sqrt x } \right]\]

Answer 1

Hint: Here, we will use the Integration by Parts formula to simplify the integrand. Then by using the suitable Integral formula, we will find the integral of the given function. Integration is defined as the summation of all the discrete data.

Formula Used:
We will use the following formula:
1. Derivative formula: \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\], \[\dfrac{d}{{dx}}\left( C \right) = 1\]
2. Integration by Parts: \[\int {uvdx = uv - \int {vdu} } \]
3. Integral Formula: \[\int {\cos tdt = \sin t} \], \[\int {\sin tdt = - \cos t} \]

Complete Step by Step Solution:
We are given an integral function \[\int {\cos \sqrt x dx} \] .
Let the given integral function be \[I\]
\[I = \int {\cos \sqrt x dx} \] …………………………………………….\[\left( 1 \right)\]
Now, we will substitute a variable for the radical expression in the integrand, we get
\[t = \sqrt x = {\left( x \right)^{\dfrac{1}{2}}}\] ……………………………………...\[\left( 2 \right)\]
Now, we will differentiate the variable with respect to \[x\] using the derivative formula \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\], we get
\[ \Rightarrow \dfrac{{dt}}{{dx}} = \dfrac{1}{2}{x^{\dfrac{1}{2} - 1}}\]
\[ \Rightarrow dt = \dfrac{1}{{2\sqrt x }}dx\]
Now, by rewriting the equation, we get
\[ \Rightarrow dx = 2\sqrt x dt\]
\[ \Rightarrow dx = 2tdt\] ……………………………………………\[\left( 3 \right)\]
Substituting the equation \[\left( 2 \right)\] and \[\left( 3 \right)\] in equation \[\left( 1 \right)\] , we get
\[I = \int {\cos t \cdot 2tdt} \]
\[ \Rightarrow I = 2\int {t\cos tdt} \]
Now, by using Integration by Parts formula \[\int {uvdx = uv - \int {vdu} } \] for the Integral function, we get \[u = t\] according to ILATE rule and \[v = \cos t\] .
Now, we will differentiate the variable \[u\], so we get
\[du = dt\]
Now, we will integrate the variable \[v\] using the formula \[\int {\cos tdt = \sin t} \], so we get
\[\int {\cos tdt = \sin t} \]
Substituting differentiated variable and integrated variable in the integration by parts formula, we get
\[ \Rightarrow \int {t\cos tdt = t\sin t - \int {\sin tdt} } \]
Now, by using the integral formula \[\int {\sin tdt = - \cos t} \], we get
\[ \Rightarrow \int {t\cos tdt = t\sin t - \left( { - \cos t} \right)} \]
\[ \Rightarrow \int {t\cos tdt = t\sin t + \cos t} \] …………………………………………\[\left( 4 \right)\]
Now, by substituting \[I = 2\int {t\cos tdt} \] in the equation \[\left( 4 \right)\], we get
\[ \Rightarrow I = 2\left[ {t\sin t + \cos t} \right] + c\]
Now, by substituting the equation \[\left( 2 \right)\], we get
\[ \Rightarrow I = 2\left[ {\sqrt x \sin \sqrt x + \cos \sqrt x } \right]\]
Therefore, the value of \[\int {\cos \sqrt x dx} \] is \[2\left[ {\sqrt x \sin \sqrt x + \cos \sqrt x } \right]\].

Thus, option (D) is the correct answer.

Note:
We know that Integration is the process of adding small parts to find the whole parts. While performing the Integration by Parts, the first function is selected according to ILATE rule where Inverse Trigonometric function, followed by Logarithmic function, Arithmetic Function, Trigonometric Function and at last Exponential Function. Integration by Parts is applicable only when the integrand is a product of two Functions.