Question

Evaluate the given integral $\int{\left( \sqrt{\dfrac{\cos x}{x}}-\sqrt{\dfrac{x}{\cos x}}\sin x \right)dx}$
$\begin{aligned} & a)-\sqrt{x\cos x}+C \\\ & b)\sqrt{x\cos x}+C \\\ & c)2\sqrt{x\cos x}+C \\\ & d)C-2\sqrt{x\cos x} \\\ \end{aligned}$

Answer 1

First we will write the expression inside integral in the form of $\dfrac{p}{q}$ by cross multiplication. Then we know that $(f.g)'=f'g+g'f$ hence we can use this formula to simplify the numerator. Then we will use a method of substitution to solve the integration.

Complete step by step answer:
Now first consider $\int{\left( \sqrt{\dfrac{\cos x}{x}}-\sqrt{\dfrac{x}{\cos x}}\sin x \right)dx}$
Let us say the value of this integral is I, then we have $I=\int{\left( \sqrt{\dfrac{\cos x}{x}}-\sqrt{\dfrac{x}{\cos x}}\sin x \right)dx}$
Rearranging the terms we get $I=\int{\left( \dfrac{\sqrt{\cos x}}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{\cos x}}\sin x \right)dx}$
Cross multiplying the terms we get,
$I=\int{\left( \dfrac{\sqrt{\cos x}.\sqrt{\cos x}-\sqrt{x}\sqrt{x}\sin x}{\sqrt{x\cos x}} \right)dx}$
Now we know that for any a $\sqrt{a}.\sqrt{a}=a$
Hence, we can write the above equation as
$I=\int{\left( \dfrac{\cos x-x\sin x}{\sqrt{x\cos x}} \right)dx}..................(1)$
Now we will try to simplify the numerator.
Consider the numerator $\cos x-x\sin x$
We know that $\dfrac{d(\cos x)}{dx}=-\sin x$ and $\dfrac{d(x)}{dx}=1$
Now substituting this in equation (1) we get
$I=\int{\left( \dfrac{\cos x\left( \dfrac{d(x)}{dx} \right)+x\left( \dfrac{d(\cos x)}{dx} \right)}{\sqrt{x\cos x}} \right)dx}.$
But we know that $f(x)g'(x)+g(x)f'(x)=\dfrac{d(f(x).g(x))}{dx}$
Hence we will use this to get
$I=\int{\left( \dfrac{\left( \dfrac{d(x.\cos x)}{dx} \right)}{\sqrt{x\cos x}} \right)dx}.$
Now cancelling dx we get from the denominator and the numerator we get

& I=\int{\left( \dfrac{\left( \dfrac{d(x.\cos x)}{1} \right)}{\sqrt{x\cos x}} \right)} \\\ & I=\int{\left( \dfrac{d(x\cos x)}{\sqrt{x\cos x}} \right)} \\\ \end{aligned}$$ Now we will solve this by method of integration. Hence, to do so let us substitute x.cosx as t. Hence we get $I=\int{\dfrac{dt}{\sqrt{t}}}$ Now we know that $\dfrac{1}{\sqrt{t}}={{t}^{-\dfrac{1}{2}}}$ Hence we have $I=\int{{{t}^{-\dfrac{1}{2}}}}dt$ Now integration of ${{x}^{n}}=\dfrac{{{x}^{n+1}}}{n+1}$+ C Hence we get $I=\int{{{t}^{-\dfrac{1}{2}}}}dt=\dfrac{{{t}^{1-\dfrac{1}{2}}}}{1-\dfrac{1}{2}}+C=2{{t}^{\dfrac{1}{2}}}+C$ Hence we have $I=2{{t}^{\dfrac{1}{2}}}+C$ Now we know that ${{t}^{\dfrac{1}{2}}}=\sqrt{t}$ hence using this we get $I=2\sqrt{t}+C$ Now let us substitute the value of t as x.cosx Hence we get $I=2\sqrt{x\cos x}+C$ **Option d is the correct option** **Note:** Note that dx in our cancels out and the variable of integration changes to t. While taking differentiation of cosx note that it is – sinx and not just sinx. Also in the end do not forget to substitute the value of t and add constant since this is an indefinite integration.