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

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

a. sin⁡(2x)4+x2+c\dfrac{\sin \left( 2x \right)}{4}+\dfrac{x}{2}+c
b. sin⁡(2x)4−x2+c\dfrac{\sin \left( 2x \right)}{4}-\dfrac{x}{2}+c
c. 2sin⁡(2x)+c2\sin \left( 2x \right)+c
d. sin⁡(x)+cos⁡(x)+c\sin \left( x \right)+\cos \left( x \right)+c

Explanation

Solution

Hint: In this question, we have to find out the indefinite integral of sin⁡(x)\sin \left( x \right) with respect to cos⁡(x)\cos \left( x \right).We know that indefinite integral of f(x) with respect to x is given by ∫f(x)dx\int{f\left( x \right)dx} in a similar way the indefinite integral of sin⁡(x)\sin \left( x \right) with respect to cos⁡(x)\cos \left( x \right) is given by
∫sin⁡(x)d(cos⁡(x))\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:
∫f(x)dg(x)\int{f\left( x \right)dg\left( x \right)}
So, indefinite integral of sin⁡(x)\sin \left( x \right) with respect to cos⁡(x)\cos \left( x \right) is
=∫sin⁡(x)d(cos⁡(x))\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 ∫f(x)dx\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:
=∫sin⁡(x)d(cos⁡(x))dxdx\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 d(cos⁡(x))dx=−sin⁡(x)\dfrac{d\left( \cos \left( x \right) \right)}{dx}=-\sin \left( x \right) .Putting the value of d(cos⁡(x))dx=−sin⁡(x)\dfrac{d\left( \cos \left( x \right) \right)}{dx}=-\sin \left( x \right)
We get,
=∫sin⁡(x)×(−sin⁡(x))×dx =∫−sin⁡2(x)dx \begin{aligned} & =\int{\sin \left( x \right)\times \left( -\sin \left( x \right) \right)\times dx} \\\ & =\int{-{{\sin }^{2}}\left( x \right)dx} \\\ \end{aligned}
We know that (cos⁡(2x)=1−2sin⁡2(x))\left( \cos \left( 2x \right)=1-2{{\sin }^{2}}\left( x \right) \right) and sin⁡2(x)=1−cos⁡(2x)2{{\sin }^{2}}\left( x \right)=\dfrac{1-\cos \left( 2x \right)}{2}
Now, putting the value of sin⁡2(x){{\sin }^{2}}\left( x \right) we get,

& =-\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{aligned}$$ We know that $\int{\cos \left( x \right)dx}=\sin \left( x \right)$ so, $\begin{aligned} & =-\dfrac{1}{2}\left( x-\dfrac{\sin \left( 2x \right)}{2} \right)+c \\\ & =\dfrac{\sin \left( 2x \right)}{4}-\dfrac{x}{2}+c \\\ \end{aligned}$ 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.