Question

What is $\int{\left( x\cos x+\sin x \right)dx}$ equal to? (C is an arbitrary constant)
[a] xsinx+C
[b] xcosx+C
[c] -xsinx+C
[d] -xcosx+C

Answer 1

Hint: Use the fact that $\int{\left( f\left( x \right)+g\left( x \right) \right)dx}=\int{f\left( x \right)dx}+\int{g\left( x \right)dx}$ and hence prove that the given integral is equal to $\int{x\cos xdx}+\int{\sin x}dx$. In the first integral use integration by parts taking x as first function and cosx as second function as per the ILATE rule. The result of the first function and second function can be related and hence find the value of the given integral.

Complete step-by-step answer:
Let $I=\int{\left( x\cos x+\sin x \right)dx}$
We know that $\int{\left( f\left( x \right)+g\left( x \right) \right)dx}=\int{f\left( x \right)dx}+\int{g\left( x \right)dx}$(Linearity of integration).
Hence, we have
$I=\int{x\cos xdx}+\int{\sin x}dx={{I}_{1}}+{{I}_{2}}$, where ${{I}_{1}}=\int{x\cos xdx}$ and ${{I}_{2}}=\int{\sin xdx}$
Evaluating ${{I}_{1}}=\int{x\cos xdx}$
We know that $\int{uvdx}=u\int{vdx}-\int{\left( \dfrac{du}{dx}\int{vdx} \right)dx}$. This is known as integration by parts rule. Here $u$ is called the first function and $v$ is known as the second function.
While solving questions using integration by parts, it is important to choose the first function and second function in such a way that the integral simplifies. A general order of preference for first function is given by ILATE rule
I = Inverse Trigonometric
L = Logarithmic
A = Algebraic
T = Trigonometric
E = Exponential.
Hence, according to the ILATE rule, we choose u(x) = x and v(x) = cosx.
We have $\int{u\left( x \right)}=\int{\cos xdx}=\sin x$ and $\dfrac{dv\left( x \right)}{dx}=\dfrac{d}{dx}\left( x \right)=1$
Hence, we have
$\int{x\cos xdx}=x\sin x-\int{\sin xdx}+C$
We have $\int{\sin xdx}={{I}_{2}}$
Hence, we have
${{I}_{1}}=x\sin x-{{I}_{2}}+C$
Adding ${{I}_{2}}$ on both sides, we get
${{I}_{1}}+{{I}_{2}}=x\sin x+C$
But we know that ${{I}_{1}}+{{I}_{2}}=I$
Hence, we have
$\begin{align}
  & I=x\sin x+C \\
 & \Rightarrow \int{\left( x\cos x+\sin x \right)dx}=x\sin x+C \\
\end{align}$

So, the correct answer is “Option A”.

Note: [1] Verification: In case the integrands are simple, we should always verify the correctness of our solution. We can verify the correctness of our solution by checking that the derivative of our answer is equal to the integrand.
We have $\begin{align}
  & \dfrac{d}{dx}\left( x\sin x+c \right)=x\dfrac{d}{dx}\sin x+\sin x\dfrac{d}{dx}x+0 \\
 & =x\cos x+\sin x \\
\end{align}$
Hence, we have
$\dfrac{d}{dx}\left( x\sin x+C \right)=x\cos x+\sin x$
Hence our solution is verified to be correct.