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# Evaluate the given Integral:$\int \left(\sqrt{\mathrm{cos}x}\right)sinxdx$$\int{(\sqrt{\cos x})sinx}dx$ Verified
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Hint: Integral expression contains a function and its derivative, hence we use substitution method to reduce the integral into standard form.

Consider the expression,
$\int \left(\sqrt{\mathrm{cos}x}\right)sinxdx$$\int{(\sqrt{\cos x})sinx}dx$
It consists of two functions, where one function i.e. $\mathrm{cos}x$$\cos x$ is the derivative of another which is $\mathrm{sin}x$$\sin x$.
So, we can use the method of integration by substitution.
In this method we substitute one of the functions to reduce the expression into standard form.
Now let us consider, $\mathrm{cos}x=t$$\cos x=t$
Differentiating both sides with respect to x, we get
We know,$\frac{d\left(\mathrm{cos}x\right)}{dx}=-\mathrm{sin}x$$\dfrac{d(\cos x)}{dx}=-\sin x$
Now,
$\dfrac{d(\cos x)}{dx}\ =\dfrac{dt}{dx}$
$-\mathrm{sin}x=\frac{dt}{dx}$$-\sin x=\dfrac{dt}{dx}$
$-\mathrm{sin}xdx=dt$$-\sin xdx=dt$
Substitute $\mathrm{sin}xdx=-dt$$\sin xdx=-dt$ in the expression, we get
$~\int{(\sqrt{\cos x})sinx}dx=-\int{(\sqrt{t}\ )dt}$
We know, $\int{x^ndx}=\ \dfrac{{{x}^{n+1}}}{n+1}+C$

So, after integrating we get
$~-\int{(\sqrt{t}\ )dt}=-\dfrac{{{t}^{\frac{3}{2}}}}{\dfrac{3}{2}}+C$
$~-\int{(\sqrt{t}\ )dt}=-\dfrac{2}{3}\ {{t}^{\frac{3}{2}}}+C$
Re-substitute the value of t in terms of x, we get
$~\int{(\sqrt{\cos x})sinx}dx=-\dfrac{2}{3}\ {{\cos }^{\frac{3}{2}}}x+C$
Note: Whenever an integrating expression consists of more than one function convert it into standard form by reduction method of integration such as substitution method of integration
Last updated date: 24th Sep 2023
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