Solveeit Logo
Login

Question

Question: Evaluate \[\int {\sin \sqrt x } dx = \] 1) \[2\left( {\sin \sqrt x - \sqrt x .\cos \sqrt x } \rig...

Evaluate ∫sin⁡xdx=\int {\sin \sqrt x } dx =

  1. 2(sin⁡x−x.cos⁡x)+c2\left( {\sin \sqrt x - \sqrt x .\cos \sqrt x } \right) + c
  2. 2(sin⁡x+x.cos⁡x)+c2\left( {\sin \sqrt x + \sqrt x .\cos \sqrt x } \right) + c
  3. 2(sin⁡x−12xcos⁡x)+c2\left( {\sin \sqrt x - \dfrac{1}{2}\sqrt x \cos \sqrt x } \right) + c
  4. (sin⁡x+12xcos⁡x)+c\left( {\sin \sqrt x + \dfrac{1}{2}\sqrt x \cos \sqrt x } \right) + c
Explanation

Solution

Hint : In order to determine the given definite integral, we have to use the substitution method of integration by substituting xxwith t. then, we use the product rule for further differentiation, the formula is ∫udv=uv−∫vdu\int {udv = uv - \int {vdu} } to evaluate to integral to obtain the required answer.

Complete step-by-step answer :
We are given the integral ∫sin⁡xdx\int {\sin \sqrt x } dx ----------(1)
Now, let us assume t=xt = \sqrt x ----------(2)
Perform the first derivative of the above assumed equation, then

⇒dtdx=12x ⇒dx=2xdt   \Rightarrow \dfrac{{dt}}{{dx}} = \dfrac{1}{{2\sqrt x }} \\\ \Rightarrow dx = 2\sqrt x dt \;

Since, t = \sqrt x $$$$ \Rightarrow dx = 2tdt ---------(3)
Here, we use the substitution method of integration to solve the above integral
As per the assumption, we have
∫2tsin⁡tdt=2∫tsin⁡tdt\int {2t\sin t} dt = 2\int {t\sin tdt}
With the the product rule ∫udv=uv−∫vdu\int {udv = uv - \int {vdu} } and trigonometric identity sin⁡t=−cos⁡t\sin t = - \cos t and ∫cos⁡t=sin⁡t\int {\cos t = \sin t} . Since, we have assumed expression x as ‘t’,
Let us assume, u=t⇒du=1⋅dtu = t \Rightarrow du = 1 \cdot dt and dv=sin⁡t⇒v=−cos⁡tdv = \sin t \Rightarrow v = - \cos t
we can write the function as
2∫tsin⁡tdt=2[t(−cos⁡t)−∫(−cos⁡t)dt]2\int {t\sin t} dt = 2\left[ {t( - \cos t) - \int {( - \cos t)dt} } \right]
Now we can simplify the integration with the trigonometric identity∫cos⁡t=sin⁡t\int {\cos t = \sin t} , then
2∫tsin⁡tdt=2(−tcos⁡t+sin⁡t)+c2\int {t\sin t} dt = 2\left( { - t\cos t + \sin t} \right) + c
We can write back the assumption value ‘t’. Since ,t=xt = \sqrt x
∫sin⁡xdx=2(−xcos⁡x+sin⁡x)+C\int {\sin \sqrt x } dx = 2\left( { - \sqrt x \cos \sqrt x + \sin \sqrt x } \right) + C
Simplifying further, we get
∫sin⁡xdx=2(sin⁡x−xcos⁡x)+C\int {\sin \sqrt x } dx = 2\left( {\sin \sqrt x - \sqrt x \cos \sqrt x } \right) + C
Where, C is constant.
Therefore, the option (1) is the correct one.
Hence, we solved the integral ∫sin⁡xdx=2(sin⁡x−xcos⁡x)+C\int {\sin \sqrt x } dx = 2\left( {\sin \sqrt x - \sqrt x \cos \sqrt x } \right) + C
So, the correct answer is “Option 1”.

Note : 1. Use a trigonometric identity formula carefully while evaluating the integrals.
2. Different types of methods of Integration:
Integration by Substitution
Integration by parts
Integration of rational algebraic function by using partial fraction
3. Integration by Substitution: The method of evaluating the integral by reducing it to standard form by a proper substitution is called integration by substitution.
4. Product rule for integration. Formula for the two functions: u and v is mentioned above.
After evaluating this integral we substitute back the value of t.
∫udv=uv−∫vdu\int {udv = uv - \int {vdu} }