Question

The integral $\int {\left(\frac{x}{x\sin x+\cos x}\right)}^{2}dx$ is equal to (where $C$ is a constant of integration)

• A. $\tan x-\left(\frac{x\sec x}{x\sin x+\cos x}\right)+C$
• B. $\sec x-\left(\frac{x\tan x}{x\sin x+\cos x}\right)+C$
• C. $\sec x+\left(\frac{x\tan x}{x\sin x+\cos x}\right)+C$
• D. $\tan x+\left(\frac{x\sec x}{x\sin x+\cos x}\right)+C$

Answer 1

The correct option is A

$\tan x-\left(\frac{x\sec x}{x\sin x+\cos x}\right)+C$

Explanation for the correct answer:

Step 1: Reduce the given integral to use integration by parts

Given integration: $\int {\left(\frac{x}{x\sin x+\cos x}\right)}^{2}dx$

$\int {\left(\frac{x}{x\sin x+\cos x}\right)}^{2}dx=\int \frac{x·x}{{\left(x\sin x+\cos x\right)}^2}dx$

$$\begin{gathered} =\int x\sec x·\frac{x\cos x}{{\left(x\sin x+\cos x\right)}^2}dx \\ =\int x\sec x·\frac{x\cos x}{{\left(x\sin x+\cos x\right)}^2}dx\left[\because \mathbf{\int }\mathit{u}\mathbf{·}\mathit{v}\mathbf{d}\mathit{x}\mathbf{=}\mathit{v}\mathbf{\int }\mathit{u}\mathbf{d}\mathit{x}\mathbf{-}\mathbf{\int }\mathbf{[}\mathbf{\left(}\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}\mathbf{\right)}\mathbf{\int }\mathbf{u}\mathbf{d}\mathbf{x}\mathbf{]}\mathbf{d}\mathit{x}\right] \\ =x\sec x\int \frac{x\cos x}{{\left(x\sin x+\cos x\right)}^2}dx-\int \left[\frac{d}{dx}\left(x\sec x\right)\int \frac{x\cos x}{{\left(x\sin x+\cos x\right)}^2}dx\right]dx...\left(i\right) \end{gathered}$$

Step 2: Solve the integrals required in (i) by substitution method

Consider the integral $\int \frac{x\cos x}{{\left(x\sin x+\cos x\right)}^2}dx$.

Let us assume that, $\left(x\sin x+\cos x\right)=t$.

Differentiate both sides of the equation.

$$\begin{gathered} d\left(x\sin x+\cos x\right)=dt \\ \Rightarrow d\left(x\sin x\right)+d\left(\cos x\right)=dt \\ \Rightarrow \sin xdx+xd\left(\sin x\right)-\sin xdx=dt \\ \Rightarrow x\cos xdx+\sin xdx-\sin xdx=dt \\ \Rightarrow x\cos xdx=dt \end{gathered}$$

$\therefore \int \frac{x\cos x}{{\left(x\sin x+\cos x\right)}^2}dx=\int \frac{dt}{t^2}$

$$\begin{gathered} =\left(-2+1\right)t^{\left(-2+1\right)} \\ =-\frac{1}{t} \\ =-\frac{1}{x\sin x+\cos x}\left[\because \left(x\sin x+\cos x\right)=t\right] \end{gathered}$$

Step 3: Resubstitute the value of these integrals in (i) to find the required solution

Thus, $\int {\left(\frac{x}{x\sin x+\cos x}\right)}^{2}dx=x\sec x\left(-\frac{1}{x\sin x+\cos x}\right)-\int \left[\left(\sec x+x\sec x·\tan x\right)\left(-\frac{1}{x\sin x+\cos x}\right)\right]dx$

$$\begin{gathered} =-\frac{x\sec x}{x\sin x+\cos x}+\int \left[\left(\frac{1}{\cos x}+x·\frac{1}{\cos x}·\frac{\sin x}{\cos x}\right)\left(\frac{1}{x\sin x+\cos x}\right)\right]dx \\ =-\frac{x\sec x}{x\sin x+\cos x}+\int \left[\left(\frac{\cos x+x\sin x}{{\cos }^{2}x}\right)\left(\frac{1}{x\sin x+\cos x}\right)\right]dx \\ =-\frac{x\sec x}{x\sin x+\cos x}+\int \left(\frac{1}{{\cos }^{2}x}\right)dx \\ =-\frac{x\sec x}{x\sin x+\cos x}+\int {\sec }^{2}xdx \\ =\tan x-\frac{x\sec x}{x\sin x+\cos x}+C \end{gathered}$$

Therefore, the integral $\int {\left(\frac{x}{x\sin x+\cos x}\right)}^{2}dx$ is equal to $\tan x-\left(\frac{x\sec x}{x\sin x+\cos x}\right)+C$, so option (A) is the correct answer.