Question

How do you integrate \[\sqrt x .(\cos \sqrt x )\]?

Answer 1

Hint: Here in this question given an indefinite integral, we have to find the integrated value of given function. it can be solved by the method of integration by parts by separating the function as \[u\]and \[v\], later integrated by using the standard formulas of integration. And by further simplification we get the required solution.

Complete step-by-step solution:
Integration by parts is a technique for performing indefinite integration or definite integration by expanding the differential of a product of functions \[d\left( {uv} \right)\] and expressing the original integral in terms of a known integral \[\int {v\,du} \]. A single integration by parts starts with
\[d(uv) = u\,dv + v\,du\]
and integrates both sides,
\[\int {d(uv)} = uv = \int {u\,dv} + \int {v\,} du.\]------(1)
Rearranging gives
\[\int {u\,} dv = uv - \int v \,du.\]---------(2)
Consider the given function \[\int {\sqrt x .\cos \sqrt x } \,\,dx\]-----(3)
substitute \[t = \sqrt x \] and \[2t\,dt = dx\]. On substituting these in the equation (3) we have
 \[ \Rightarrow \int {t.\cos t} \,\,(2{t^2}\,dt)\]
on simplifying we get
\[ \Rightarrow \int {2{t^3}.\cos t} \,\,dt\]
Given integral which is not having any upper and lower limit then it’s an indefinite integral. Hence we add the C while integrating. Where, C is an arbitrary constant called as the constant of integration.\[\]
 \[u = 2{t^3}\] \[ \Rightarrow \,\,\dfrac{{du}}{{dt}} = 3{t^2}\] and
\[dv = \cos t\,dt\] \[ \Rightarrow \,\,v = - \sin t\]
Then by the method of integration by parts i.e., by the equation (2)
\[\, \Rightarrow \,\,\,\int {2{t^3}.\cos t\,dt} = \,\,2{t^3}(\sin t) - \int {\sin t(6{t^2})\,} dt\]
\[\, \Rightarrow \,\,\,\int {2{t^3}\cos t\,dt} = \,\,2{t^3}(\sin t) - \int {6{t^2}.(\sin t)\,} dt\]
again on integrating we have
 \[\, \Rightarrow \,\,\,\int {2{t^3}\cos t\,dt} = \,\,2{t^3}\,\sin t - 6{t^2}( - \cos t) - \int {12t( - \cos t)\,dt} \]
On simplifying we get
\[\, \Rightarrow \,\,\,\int {2{t^3}\cos t\,dt} = \,\,2{t^3}\,\sin t + 6{t^2}\cos t + \int {12t\,\cos t\,dt} \]
On integrating the third term of RHS, we get
\[\, \Rightarrow \,\,\,\int {2{t^3}\cos t\,dt} = \,\,2{t^3}\,\sin t + 6{t^2}\cos t + 12t\,(\sin t) + C\]
Where C is an integrating constant.
Re-substitute the value of t we have
\[\, \Rightarrow \,\,\,\int {\sqrt x \cos \sqrt x \,dx} = \,\,2{(\sqrt x )^3}\,\sin \sqrt x + 6x\cos \sqrt x + 12\sqrt x \,(\sin \sqrt x ) + C\]

Hence, the value of \[\int {\sqrt x \cos \sqrt x } \,\,dx\] is \[\,2{(\sqrt x )^3}\,\sin \sqrt x + 6x\cos \sqrt x + 12\sqrt x \,(\sin \sqrt x ) + C\]

Note: While integrating the trigonometric functions, we simplify the trigonometric functions as much as possible by using the trigonometry ratios or by trigonometry identities. The integration by substitution is the easiest way to integrate. The function and its derivative must be present while substituting.