Answer

Verified

342k+ views

**Hint:**The given integral $\sin \left( {{x}^{\dfrac{1}{2}}} \right)dx$ has \[{{x}^{\dfrac{1}{2}}}\] or \[\sqrt{x}\] as an argument to the sine function, which is making it complex. So we will simplify the integral by substituting \[{{x}^{\dfrac{1}{2}}}\] equal to some variable, say $t$. On making this substitution our integral will become simplified. Then we have to use the by-parts method to solve the integral obtained. Finally, we have to back substitute $t$ to \[{{x}^{\dfrac{1}{2}}}\] to get the final integral.

**Complete step-by-step answer:**

The integral given in the above question is

\[I=\int{\sin \left( {{x}^{\dfrac{1}{2}}} \right)dx}..........(i)\]

As can be seen above, the square root function \[{{x}^{\dfrac{1}{2}}}\] as an argument to the sine function is making the integral complex. So we first have to simplify the above integral by removing the square root function by substituting it to some variable $t$, that is,

$\Rightarrow t={{x}^{\dfrac{1}{2}}}.........(ii)$

Differentiating both sides with respect to $x$, we have

$\Rightarrow \dfrac{dt}{dx}=\dfrac{d\left( {{x}^{\dfrac{1}{2}}} \right)}{dx}$

Now, we know that the differentiation of ${{x}^{n}}$ is equal to $n{{x}^{n-1}}$. So the above equation becomes

\[\begin{align}

& \Rightarrow \dfrac{dt}{dx}=\dfrac{1}{2}{{x}^{\dfrac{1}{2}-1}} \\

& \Rightarrow \dfrac{dt}{dx}=\dfrac{1}{2}{{x}^{-\dfrac{1}{2}}} \\

& \Rightarrow \dfrac{dt}{dx}=\dfrac{1}{2{{x}^{\dfrac{1}{2}}}} \\

\end{align}\]

Substituting (ii) in the above equation, we get

\[\Rightarrow \dfrac{dt}{dx}=\dfrac{1}{2t}\]

By cross multiplying, we can write

\[\begin{align}

& \Rightarrow 2tdt=dx \\

& \Rightarrow dx=2tdt.........(iii) \\

\end{align}\]

Substituting (ii) and (iii) in (i), we get

$\begin{align}

& \Rightarrow I=\int{\sin t\left( 2tdt \right)} \\

& \Rightarrow I=\int{2t\sin tdt}......(iv) \\

\end{align}$

Now, we will use the by parts method to solve the above integral. From the by parts method, we know that

$\int{f\left( t \right)g\left( t \right)dt}=f\left( t \right)\int{g\left( t \right)dt}-\int{f'\left( t \right)\left( \int{g\left( t \right)dt} \right)dt}$

Choosing $f\left( t \right)=2t$ and $g\left( t \right)=\sin t$, the integral in (iv) can be written as

\[\begin{align}

& \Rightarrow I=2t\int{\sin tdt}-\int{\dfrac{d\left( 2t \right)}{dt}\left( \int{\sin tdt} \right)dt} \\

& \Rightarrow I=2t\int{\sin tdt}-\int{2\left( \int{\sin tdt} \right)dt} \\

\end{align}\]

We know that \[\int{\sin tdt}=-\cos t\]. Putting it in the above integral, we get

\[\begin{align}

& \Rightarrow I=2t\left( -\cos t \right)-\int{2\left( -\cos t \right)dt} \\

& \Rightarrow I=-2t\cos t+2\int{\cos tdt} \\

\end{align}\]

Now, we know that \[\int{\cos tdt}=\sin t\]. Putting it above, we get

\[\begin{align}

& \Rightarrow I=-2t\cos t+2\sin t+C \\

& \Rightarrow I=2\sin t-2t\cos t+C \\

\end{align}\]

Finally, substituting (ii) in the above equation, we get

\[\begin{align}

& \Rightarrow I=2\sin {{x}^{\dfrac{1}{2}}}-2{{x}^{\dfrac{1}{2}}}\cos {{x}^{\dfrac{1}{2}}}+C \\

& \Rightarrow I=2\left( \sin {{x}^{\dfrac{1}{2}}}-{{x}^{\dfrac{1}{2}}}\cos {{x}^{\dfrac{1}{2}}} \right)+C \\

\end{align}\]

Hence, the integral of $\sin \left( {{x}^{\dfrac{1}{2}}} \right)dx$ is equal to \[2\left( \sin {{x}^{\dfrac{1}{2}}}-{{x}^{\dfrac{1}{2}}}\cos {{x}^{\dfrac{1}{2}}} \right)+C\].

**Note:**Do not forget to add a constant after the integration since we have solved an indefinite integral. Also, while substituting ${{x}^{\dfrac{1}{2}}}=t$, do not replace $dx$ by $dt$ directly. We have to take the differential on both sides of the equation ${{x}^{\dfrac{1}{2}}}=t$ for obtaining $dt$ in terms of $dx$.

Recently Updated Pages

Basicity of sulphurous acid and sulphuric acid are

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

What is the stopping potential when the metal with class 12 physics JEE_Main

The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main

Using the following information to help you answer class 12 chemistry CBSE

Why should electric field lines never cross each other class 12 physics CBSE

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

What is the color of ferrous sulphate crystals? How does this color change after heating? Name the products formed on strongly heating ferrous sulphate crystals. What type of chemical reaction occurs in this type of change.

Write the difference between soap and detergent class 10 chemistry CBSE

Give 10 examples of unisexual and bisexual flowers

Differentiate between calcination and roasting class 11 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is the difference between anaerobic aerobic respiration class 10 biology CBSE

a Why did Mendel choose pea plants for his experiments class 10 biology CBSE