Question

The indefinite integral of $\sin \left( x \right)$ w. r .t $\cos \left( x \right)$ is

a. $\dfrac{\sin \left( 2x \right)}{4}+\dfrac{x}{2}+c$
b. $\dfrac{\sin \left( 2x \right)}{4}-\dfrac{x}{2}+c$
c. $2\sin \left( 2x \right)+c$
d. $\sin \left( x \right)+\cos \left( x \right)+c$

Answer 1

Hint: In this question, we have to find out the indefinite integral of $\sin \left( x \right)$ with respect to $\cos \left( x \right)$.We know that indefinite integral of f(x) with respect to x is given by $\int{f\left( x \right)dx}$ in a similar way the indefinite integral of $\sin \left( x \right)$ with respect to $\cos \left( x \right)$ is given by
 $\int{\sin \left( x \right)d(\cos (x))}$

Complete step-by-step answer:

The indefinite integral of function f(x) with respect to g(x) is defined as:
 $\int{f\left( x \right)dg\left( x \right)}$
So, indefinite integral of $\sin \left( x \right)$ with respect to $\cos \left( x \right)$ is
=$\int{\sin \left( x \right)d\left( \cos \left( x \right) \right)}$
To solve this above integral. We have to convert the above equation into $\int{f\left( x \right)dx}$ this form. For that firstly we have to find the differentiation of cos(x).
So,
We can re-write the above equation as:
=$\int{\sin \left( x \right)}\dfrac{d\left( \cos \left( x \right) \right)}{dx}dx$ (multiplying numerator and denominator by dx)
We know that the $\dfrac{d\left( \cos \left( x \right) \right)}{dx}=-\sin \left( x \right)$ .Putting the value of $\dfrac{d\left( \cos \left( x \right) \right)}{dx}=-\sin \left( x \right)$
We get,
$\begin{align}
  & =\int{\sin \left( x \right)\times \left( -\sin \left( x \right) \right)\times dx} \\
 & =\int{-{{\sin }^{2}}\left( x \right)dx} \\
\end{align}$
We know that $\left( \cos \left( 2x \right)=1-2{{\sin }^{2}}\left( x \right) \right)$ and ${{\sin }^{2}}\left( x \right)=\dfrac{1-\cos \left( 2x \right)}{2}$
Now, putting the value of ${{\sin }^{2}}\left( x \right)$ we get,
\[\begin{align}
  & =-\int{\left( \dfrac{1-\cos \left( 2x \right)}{2} \right)}dx \\
 & =-\dfrac{1}{2}\int{\left( 1-\cos \left( 2x \right) \right)dx} \\
 & =-\dfrac{1}{2}\left( \int{1dx-\int{\cos \left( 2x \right)dx}} \right) \\
\end{align}\]
We know that $\int{\cos \left( x \right)dx}=\sin \left( x \right)$ so,
$\begin{align}
  & =-\dfrac{1}{2}\left( x-\dfrac{\sin \left( 2x \right)}{2} \right)+c \\
 & =\dfrac{\sin \left( 2x \right)}{4}-\dfrac{x}{2}+c \\
\end{align}$
Hence, the correct option is option (B)
Hence, the indefinite integral of sin(x) with respect to cos(x) is $\dfrac{\sin \left( 2x \right)}{4}-\dfrac{x}{2}+c$

Note: One can also solve this question by taking sin(x) as t and using relation $\left( {{\sin }^{2}}\left( x \right)+{{\cos }^{2}}\left( x \right)=1 \right)$ to find the value of cos(x), $\left( i.e,\cos \left( x \right)=\sqrt{1-{{t}^{2}}} \right)$
Then, using the above formula
The indefinite integral of sin(x) with respect to cos(x) is
$=\int{\sin \left( x \right)d\left( \cos \left( x \right) \right)}$
Now putting the value of sin(x) and cos(x), we get
$=\int{t\times d\sqrt{1-{{t}^{2}}}}$ --(1)
After solving equation (1) you will arrive at the same answer, but this method involves a lot of calculation. To avoid such problems we can use the first method to solve this type of problem.