Question

Evaluate the given Integral:
$\int{(\sqrt{\cos x})sinx}dx$

Answer 1

Hint: Integral expression contains a function and its derivative, hence we use substitution method to reduce the integral into standard form.

Consider the expression,
$\int{(\sqrt{\cos x})sinx}dx$
It consists of two functions, where one function i.e. $\cos x$ is the derivative of another which is $\sin x$.
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, $\cos x=t$
Differentiating both sides with respect to x, we get
We know,$\dfrac{d(\cos x)}{dx}=-\sin x$
Now,
$\dfrac{d(\cos x)}{dx}\ =\dfrac{dt}{dx}$ 
$-\sin x=\dfrac{dt}{dx}$
$-\sin xdx=dt$
Substitute $\sin xdx=-dt$ 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