Question

Evaluate the given integral: $\int{\cos \sqrt{x}\text{ }dx}$ .
(a)$\sqrt{x}\sin \sqrt{x}+\cos \sqrt{x}+c$
(b)$2\left( \sin \sqrt{x}-\cos \sqrt{x} \right)+c$
(c)$2\left( \sqrt{x}\sin \sqrt{x}+\sqrt{x}\cos \sqrt{x} \right)+c$
(d)$2\left( \sqrt{x}\sin \sqrt{x}+\cos \sqrt{x} \right)+c$

Answer 1

Hint: Start by letting $\sqrt{x}$ be t and solve the integral. Finally use the rule of integration by parts to reach the answer.

Complete step-by-step answer:
Let us start with the integral given in the above question.
$\int{\cos \sqrt{x}\text{ }dx}$
Now to convert the integral to a form from where we can directly integrate it, we let $\sqrt{x}$ to be t.
 $\sqrt{x}=t$
Now if we differentiate both the sides of the equation, we get
$\dfrac{1}{2\sqrt{x}}dx=dt$
$\Rightarrow \dfrac{1}{2t}dx=dt$
$\Rightarrow dx=2tdt$
So, if we substitute the terms in the integral in terms of t, our integral becomes:
\[\int{2t\cos tdt}\]
Now according to the rule of the integration by parts:
$\int{uvdx=u\int{v}}dx-\int{\left( \dfrac{du}{dx}\int{vdx} \right)}$ . So, our integral becomes:
 \[2t\int{\cos tdt}-\int{\left( \dfrac{d\left( 2t \right)}{dt}\int{\cos tdt} \right)}\]
Now we know that $\int{\cos x}dx=\operatorname{sinx}+c$ . So, our integral comes out to be:
\[2t\operatorname{sint}-\int{\left( 2\operatorname{sint} \right)dt}\]
We also know that $\int{sinx}dx=-\operatorname{cosx}+c$ . So, our integral comes out to be:
\[2t\operatorname{sint}+2cost+c\]
Now we will substitute the value of t as assumed by us to convert our answer in terms of x.
\[2\sqrt{x}\sin \sqrt{x}+2cos\sqrt{x}+c\]
\[=2\left( \sqrt{x}\sin \sqrt{x}+cos\sqrt{x} \right)+c\]
So, we can say that the value of $\int{\cos \sqrt{x}\text{ }dx}$ is equal to \[2\left( \sqrt{x}\sin \sqrt{x}+cos\sqrt{x} \right)+c\] . Hence, the answer to the above question is option (d).

Note: Don’t forget to substitute the assumed variable in the integrated expression to reach the final answer. Also, we need to remember all the basic formulas that we use for indefinite integrals as they are used in definite integrations as well.