Question

What is the antiderivative of $\dfrac{x\cos x}{{{\left( x\sin x+\cos x \right)}^{2}}}$ ?

Answer 1

Hint: An antiderivative of a function f is a differentiable function F whose derivative is equal to the original function f. This can be stated symbolically as F' = f. Now, simply we need to do the substitution and will get the answer.

Complete step-by-step solution:
We say that the process of solving for antiderivatives is called antidifferentiation (or indefinite integration), and its opposite operation is called differentiation, which is the process of finding a derivative. Antiderivatives are often denoted by capital Roman letters such as F and G.
We also know that antiderivative can also be used to compute the definite integrals by using the fundamental theorems of calculus. Antiderivative is commonly asked as to evaluate the indefinite integral of any function.
We see that antiderivatives of elementary function are considerably harder than just evaluating their derivatives. We know and can explore many properties of antiderivatives in order to solve quickly and get handy with these. The linearity of integration basically removes the complex integrand into simpler ones and hence it becomes easy for us to evaluate majorly in the case of definite integrals where we are provided with the limits that we need to substitute at the place of the dependent variable and hence get the definite answer.
Now in the question we need to find the antiderivative of $\dfrac{x\cos x}{{{\left( x\sin x+\cos x \right)}^{2}}}$.
So, \[I=\int{\dfrac{x\cos x}{{{\left( x\sin x+\cos x \right)}^{2}}}dx}\]
Now, in order to integrate this,
let $t=x\sin x+\cos x$
now, differentiating this we get,
$\begin{align}
  & dt=\left( x\cos x+\sin x-\sin x \right)dx \\
 & \Rightarrow dt=x\cos xdx \\
\end{align}$
Now, our integral becomes
 \[\begin{align}
  & I=\int{\dfrac{x\cos x}{{{t}^{2}}x\cos x}dt} \\
 & \Rightarrow I=\int{\dfrac{1}{{{t}^{2}}}dt} \\
\end{align}\]
Now, we know that the integration of $\int{{{x}^{n}}=\dfrac{{{x}^{n+1}}}{n+1}}$ where n is any integer. Therefore, we get $I=\dfrac{-1}{t}+C$ .
Now, substituting back the values of t we get, $I=\dfrac{-1}{x\sin x+\cos x}+C$
So, the integration of the given function is $I=\dfrac{-1}{x\sin x+\cos x}+C$ where C is an arbitrary constant.

Note: Directly use the inverse trigonometric properties and get the antiderivative. Use the direct integration formulas that are mentioned as identity in this section and then we can make the calculation fast by just substituting the values comparing that result.