Question

Evaluate the following integral
$\begin{aligned} & \int{\dfrac{{{x}^{2}}dx}{{{\left( x\sin x+\cos x \right)}^{2}}}} \\\ & [a]\ \dfrac{\sin x+\cos x}{x\sin x+\cos x} \\\ & [b]\ \dfrac{x\sin x-\cos x}{x\sin x+\cos x} \\\ & [c]\ \dfrac{\sin x-\cos x}{x\sin x+\cos x} \\\ & [d]\text{ None of these} \\\ \end{aligned}$

Answer 1

Hint: Observe that $\dfrac{d}{dx}\left( x\sin x+\cos x \right)=x\cos x+\sin x-\sin x=x\cos x$.So we want to convert our integral such that it has xcosxdx in the numerator. Since we know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1,$ we have ${{x}^{2}}=x\sin x\left( x\sin x+\cos x \right)+x\cos x\left( -\sin x+x\cos x \right)$
Hence prove that the given integral is equal to $\int{\dfrac{x\sin x\left( x\sin x+\cos x \right)}{{{\left( x\sin x+\cos x \right)}^{2}}}dx-}\int{\dfrac{x\cos x\left( x\cos x-\sin x \right)}{{{\left( x\sin x+\cos x \right)}^{2}}}dx}$. In the second integral use integration by parts taking $\dfrac{x\cos x}{{{\left( x\sin x+\cos x \right)}^{2}}}$ as first function(Observe that this is done since the derivative of xsinx+cosx is equal to the numerator) and $\left( x\cos x-\sin x \right)$ as second function. Observe that the latter integral is produced on integration by parts in the negative of the first integral and hence they cancel each other. Hence find the value of the given integral. Verify your answer.

Complete step by step solution:
Let $I=\int{\dfrac{{{x}^{2}}dx}{{{\left( x\sin x+\cos x \right)}^{2}}}}$
Now, consider the expression $E=x\sin x\left( x\sin x+\cos x \right)+x\cos x\left( x\cos x-\sin x \right)$
Expanding the terms, we get
$E={{x}^{2}}{{\sin }^{2}}x+x\sin x\cos x+{{x}^{2}}{{\cos }^{2}}x-x\sin x\cos x={{x}^{2}}\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)$
we know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$
Hence, we have
$E={{x}^{2}}$
Replacing ${{x}^{2}}$ of the integral I by E, we get
$I=\int{\dfrac{x\sin x\left( x\sin x+\cos x \right)+x\cos x\left( x\cos x-\sin x \right)}{{{\left( x\sin x+\cos x \right)}^{2}}}}dx$
We know that $\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}$
Hence, we have
$I=\int{\left( \dfrac{x\sin x\left( x\sin x+\cos x \right)}{{{\left( x\sin x+\cos x \right)}^{2}}}+\dfrac{x\cos x\left( x\cos x-\sin x \right)}{{{\left( x\sin x+\cos x \right)}^{2}}} \right)dx}$
Simplifying, we get
$I=\int{\left( \dfrac{x\sin x}{x\sin x+\cos x}+\dfrac{x\cos x\left( x\cos x-\sin x \right)}{{{\left( x\sin x+\cos x \right)}^{2}}} \right)dx}$
Using linearity of integration, i.e. $\int{\left( f+g \right)dx=\int{fdx}+\int{gdx}}$, we get
$\begin{aligned} & I=\int{\dfrac{x\sin x\left( x\sin x+\cos x \right)dx}{{{\left( x\sin x+\cos x \right)}^{2}}}}+\int{\dfrac{x\cos x\left( x\cos x-\sin x \right)}{{{\left( x\sin x+\cos x \right)}^{2}}}dx} \\\ & =\int{\dfrac{x\sin x}{x\sin x+\cos x}dx+\int{\dfrac{x\cos x\left( x\cos x-\sin x \right)}{{{\left( x\sin x+\cos x \right)}^{2}}}dx}} \\\ \end{aligned}$
We know that if $\dfrac{d}{dx}g\left( x \right)=v\left( x \right)$ and $\int{f\left( x \right)}dx=u\left( x \right)$, then $\int{f\left( x \right)g\left( x \right)dx}=g\left( x \right)u\left( x \right)-\int{v\left( x \right)u\left( x \right)dx}$. This is known as integration by parts rule. f(x) is known as the second function, and g(x) is known as the first function.
Taking $f\left( x \right)=\dfrac{x\cos x}{{{\left( x\sin x+\cos x \right)}^{2}}}$ and $g\left( x \right)=\left( x\cos x-\sin x \right)$, we have
$u\left( x \right)=\int{\dfrac{x\cos x}{{{\left( x\sin x+\cos x \right)}^{2}}}dx}$
In the expression for u(x), put xsinx+cosx = t
Differentiating both sides, we get
dt = (xcosx+sinx-sinx)dx = xcosxdx
Hence, we have
$u\left( x \right)=\int{\dfrac{dt}{{{t}^{2}}}}$
We know that
$\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C,n\ne -1$
Hence, we have
$u\left( x \right)=\dfrac{{{t}^{-2+1}}}{-2+1}=\dfrac{-1}{t}$
Reverting to original variable, we get
$u\left( x \right)=-\dfrac{1}{x\sin x+\cos x}$
Also, we have
$v\left( x \right)=\dfrac{d}{dx}\left( x\cos x-\sin x \right)=-x\sin x+\cos x-\cos x=-x\sin x$
Hence, by integration by parts rule, we have
$\begin{aligned} & I=\int{\dfrac{x\sin x}{{{\left( x\cos x+\sin x \right)}^{2}}}dx+u\left( x \right)g\left( x \right)-\int{v\left( x \right)u\left( x \right)dx}} \\\ & =\int{\dfrac{x\sin xdx}{{{\left( x\cos x+\sin x \right)}^{2}}}+\left( x\cos x-\sin x \right)\left( \dfrac{-1}{x\sin x+\cos x} \right)-\int{\dfrac{-x\sin x}{-\left( x\cos x+\sin x \right)}dx}} \\\ \end{aligned}$
Simplifying, we get
$I=\int{\dfrac{x\sin xdx}{{{\left( x\cos x+\sin x \right)}^{2}}}-\dfrac{x\cos x-\sin x}{x\sin x+\cos x}-\int{\dfrac{x\sin xdx}{{{\left( x\cos x+\sin x \right)}^{2}}}}}$
The first and the last integrals are the same and hence cancel each other. Hence, we have
$I=-\dfrac{x\cos x-\sin x}{x\sin x+\cos x}+C$, where C is the constant of integration.
Hence option [d] is correct as none of the options [a], [b] and [c] are correct.
Note: Verification:
We have
$\dfrac{dI}{dx}=-\dfrac{d}{dx}\left( \dfrac{x\cos x-\sin x}{x\sin x+\cos x} \right)$
Using the fact that $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$
Hene, we get
$\dfrac{dI}{dx}=-\dfrac{1}{{{\left( x\sin x+\cos x \right)}^{2}}}\left( \left( x\sin x+\cos x \right)\left( -x\sin x \right)-\left( x\cos x-\sin x \right)\left( x\cos x \right) \right)$
Simplifying, we get
$\dfrac{dI}{dx}=\dfrac{-1}{{{\left( x\sin x+\cos x \right)}^{2}}}\left( -{{x}^{2}}{{\sin }^{2}}x-x\sin x\cos x-{{x}^{2}}{{\cos }^{2}}x+x\cos x\sin x \right)$
Hence, we have
$\dfrac{dI}{dx}=\dfrac{{{x}^{2}}\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)}{{{\left( x\sin x+\cos x \right)}^{2}}}=\dfrac{{{x}^{2}}}{{{\left( x\sin x+\cos x \right)}^{2}}}$
Hence, our answer is verified to be correct.