Question

Evaluate the following integral $\int{\dfrac{{{x}^{2}}}{{{\left( x\sin x+\cos x \right)}^{2}}}dx}$

Answer 1

We here have been given the integral $\int{\dfrac{{{x}^{2}}}{{{\left( x\sin x+\cos x \right)}^{2}}}dx}$ that we need to evaluate. For this, we will first see if there is any relationship between the numerator and the denominator by differentiating the denominator of a part of it. Then we will see that we can convert this into the product of two functions such that we can calculate the integral of one of those functions. Then we will use integration by parts which is given as:
$\int{f\left( x \right)g\left( x \right)dx=f\left( x \right)}\left[ \int{g\left( x \right)dx} \right]-\int{\left( \int{g\left( x \right)dx} \right)\left( \dfrac{d}{dx}\left( f\left( x \right) \right) \right)dx}$
Here, we will give g(x) the value of the function whose integral we can calculate and f(x) to the other function. Hence, using integration by parts, we will get the value of the required integral.

Complete step by step answer:
Here, we have been asked to evaluate the limit $\int{\dfrac{{{x}^{2}}}{{{\left( x\sin x+\cos x \right)}^{2}}}dx}$. For this, we will first see if the numerator is the differential of the denominator.
The denominator is: ${{\left( x\sin x+\cos x \right)}^{2}}$
We will check for it without any power. Thus, we get:
$x\sin x+\cos x$
Now, differentiating and applying the product rule, we get:
$\begin{aligned} & \dfrac{d}{dx}\left( x\sin +\cos x \right) \\\ & \Rightarrow x\cos x+1.\sin x-\sin x \\\ & \Rightarrow x\cos x \\\ \end{aligned}$
Hence, if we divide the numerator by x and multiply it by cosx, it will become the differential of the denominator.
Thus, multiplying and dividing the numerator by x and cosx we get:

& \int{\dfrac{{{x}^{2}}}{{{\left( x\sin x+\cos x \right)}^{2}}}dx} \\\ & \Rightarrow \int{\dfrac{{{x}^{2}}\times \dfrac{x}{x}\times \dfrac{\cos x}{\cos x}}{{{\left( x\sin x+\cos x \right)}^{2}}}dx} \\\ & \Rightarrow \int{\dfrac{x}{\cos x}.\dfrac{x\cos x}{{{\left( x\sin x+\cos x \right)}^{2}}}dx} \\\ \end{aligned}$$ Now, we can see that there are two functions $f\left( x \right)=\dfrac{x}{\cos x}$ and $g\left( x \right)=\dfrac{x\cos x}{{{\left( x\sin x+\cos x \right)}^{2}}}$ in this integral. Thus, we can use integration by parts to solve this question. Now, we know that the integration by parts is done as: $\int{f\left( x \right)g\left( x \right)dx=f\left( x \right)}\left[ \int{g\left( x \right)dx} \right]-\int{\left( \int{g\left( x \right)dx} \right)\left( \dfrac{d}{dx}\left( f\left( x \right) \right) \right)dx}$ Hence, if we assume the given integral to be ‘I’, we get the required value of I as: $\begin{aligned} & I=\int{\dfrac{x}{\cos x}.\dfrac{x\cos x}{{{\left( x\sin x+\cos x \right)}^{2}}}dx} \\\ & \Rightarrow I=\dfrac{x}{\cos x}\left[ \int{\dfrac{x\cos x}{{{\left( x\sin x+\cos x \right)}^{2}}}dx} \right]-\int{\left( \int{\dfrac{x\cos x}{{{\left( x\sin x+\cos x \right)}^{2}}}dx} \right)\left( \dfrac{d}{dx}\left( \dfrac{x}{\cos x} \right) \right)dx} \\\ \end{aligned}$ Now, here we have to first calculate the value of $\int{\dfrac{x\cos x}{{{\left( x\sin x+\cos x \right)}^{2}}}dx}$. For this, let us assume xsinx+cosx=t. Thus, differentiating both sides we get: $\begin{aligned} & x\sin x+\cos x=t \\\ & \Rightarrow \left( x\cos x+\sin x-\sin x \right)dx=dt \\\ & \Rightarrow x\cos xdx=dt \\\ \end{aligned}$ Thus, putting the values of t and dt in the integral, we get: $\begin{aligned} & \int{\dfrac{x\cos x}{{{\left( x\sin x+\cos x \right)}^{2}}}dx} \\\ & \Rightarrow \int{\dfrac{dt}{{{\left( t \right)}^{2}}}} \\\ & \Rightarrow \int{\dfrac{dt}{{{t}^{2}}}} \\\ \end{aligned}$ Thus, integrating it we get: $\begin{aligned} & \int{\dfrac{dt}{{{t}^{2}}}} \\\ & \Rightarrow \int{{{t}^{-2}}dt} \\\ & \Rightarrow \dfrac{{{t}^{-2+1}}}{-2+1} \\\ & \Rightarrow \dfrac{{{t}^{-1}}}{-1} \\\ & \Rightarrow -\dfrac{1}{t} \\\ \end{aligned}$ Now, putting the value of t back in the integral, we get: $\begin{aligned} & -\dfrac{1}{t} \\\ & \Rightarrow -\dfrac{1}{x\sin x+\cos x} \\\ \end{aligned}$ Thus, the value of $\int{\dfrac{x\cos x}{{{\left( x\sin x+\cos x \right)}^{2}}}dx}$ is $-\dfrac{1}{x\sin x+\cos x}$ Now, putting this in the value of I, we get: $\begin{aligned} & I=\dfrac{x}{\cos x}\left[ \int{\dfrac{x\cos x}{{{\left( x\sin x+\cos x \right)}^{2}}}dx} \right]-\int{\left( \int{\dfrac{x\cos x}{{{\left( x\sin x+\cos x \right)}^{2}}}dx} \right)\left( \dfrac{d}{dx}\left( \dfrac{x}{\cos x} \right) \right)dx} \\\ & \Rightarrow I=\dfrac{x}{\cos x}\left( -\dfrac{1}{x\sin x+\cos x} \right)-\int{\left( -\dfrac{1}{x\sin x+\cos x} \right)\left( \dfrac{d}{dx}\left( \dfrac{x}{\cos x} \right) \right)dx} \\\ & \Rightarrow I=-\left( \dfrac{x}{\cos x(x\sin x+\cos x)} \right)+\int{\left( \dfrac{1}{x\sin x+\cos x} \right)\left( \dfrac{d}{dx}\left( \dfrac{x}{\cos x} \right) \right)dx} \\\ \end{aligned}$ Now, for the next step, we need to find the value of the differential of $\dfrac{x}{\cos x}$. Thus, differentiating it using quotient rule we get: $\begin{aligned} & \dfrac{d}{dx}\left( \dfrac{x}{\cos x} \right) \\\ & \Rightarrow \dfrac{\cos x\left( \dfrac{d}{dx}\left( x \right) \right)-x\left( \dfrac{d}{dx}\left( \cos x \right) \right)}{{{\left( \cos x \right)}^{2}}} \\\ & \Rightarrow \dfrac{\cos x.1-x\left( -\sin x \right)}{{{\cos }^{2}}x} \\\ & \Rightarrow \dfrac{\cos x+x\sin x}{{{\cos }^{2}}x} \\\ \end{aligned}$ Now, keeping this value in I, we get: $\begin{aligned} & I=-\left( \dfrac{x}{\cos x(x\sin x+\cos x)} \right)+\int{\left( \dfrac{1}{x\sin x+\cos x} \right)\left( \dfrac{d}{dx}\left( \dfrac{x}{\cos x} \right) \right)dx} \\\ & \Rightarrow I=-\dfrac{x}{\cos x\left( x\sin x+\cos x \right)}+\int{\left( \dfrac{1}{x\sin x+\cos x} \right)\left( \dfrac{\cos x+x\sin x}{{{\cos }^{2}}x} \right)dx} \\\ & \Rightarrow I=-\dfrac{x}{\cos x\left( x\sin x+\cos x \right)}+\int{\dfrac{1}{{{\cos }^{2}}x}dx} \\\ & \Rightarrow I=-\dfrac{x}{\cos x\left( x\sin x+\cos x \right)}+\int{{{\sec }^{2}}xdx} \\\ \end{aligned}$ Now, we know that $\int{{{\sec }^{2}}xdx=\tan x}$ Thus, we get the value of I as: $\begin{aligned} & I=-\dfrac{x}{\cos x\left( x\sin x+\cos x \right)}+\int{{{\sec }^{2}}xdx} \\\ & \Rightarrow I=-\dfrac{x}{\cos x\left( x\sin x+\cos x \right)}+\tan x+C \\\ \end{aligned}$ Here, C is the constant of integration. Now, simplifying I we get: $\begin{aligned} & I=-\dfrac{x}{\cos x\left( x\sin x+\cos x \right)}+\tan x+C \\\ & \Rightarrow I=-\dfrac{x}{\cos x\left( x\sin x+\cos x \right)}+\dfrac{\sin x}{\cos x}+C \\\ & \Rightarrow I=\dfrac{-x+x{{\sin }^{2}}x+\sin x\cos x}{\cos x\left( x\sin x+\cos x \right)}+C \\\ & \Rightarrow I=\dfrac{x\left( {{\sin }^{2}}x-1 \right)+\sin x\cos x}{\cos x\left( x\sin x+\cos x \right)}+C \\\ & \Rightarrow I=\dfrac{-x{{\cos }^{2}}x+\sin x\cos x}{\cos x\left( x\sin x+\cos x \right)}+C \\\ \end{aligned}$ Now, taking cosx common we get: $\begin{aligned} & I=\dfrac{-x{{\cos }^{2}}x+\sin x\cos x}{\cos x\left( x\sin x+\cos x \right)}+C \\\ & \Rightarrow I=\dfrac{\cos x\left( \sin x-x\cos x \right)}{\cos x\left( x\sin x+\cos x \right)}+C \\\ & \Rightarrow I=\dfrac{\sin x-x\cos x}{x\sin x+\cos x}+C \\\ \end{aligned}$ **Thus, the value of the given integral is $\dfrac{\sin x-x\cos x}{x\sin x+\cos x}+C$.** **Note:** We here can see that we have only added the constant of integration in the final answer even though we did multiple integrations in itself. This is because multiple integrations will lead to multiple constants of integration and it will create confusion even though at the end all constants combine to form one single constant. Hence, to avoid all this we only add a constant at the last.