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Question: How do you find the antiderivative of \( \sin x\cos x \) ?...

How do you find the antiderivative of sin⁡xcos⁡x\sin x\cos x ?

Explanation

Solution

Hint : Derivative means finding a very small part of a whole quantity, so as the name suggests antiderivative means finding the whole quantity from a given small part. Thus we have to find the integration of the given quantity. The function which has to be integrated is a trigonometric function, so we can simplify it by using trigonometric identities and then find its integration.

Complete step-by-step answer :
We have to find ∫sin⁡xcos⁡x\int {\sin x\cos x}
We know that,
2sin⁡xcos⁡x=sin⁡2x ⇒sin⁡xcos⁡x=12×sin⁡2x   2\sin x\cos x = \sin 2x \\\ \Rightarrow \sin x\cos x = \dfrac{1}{2} \times \sin 2x \;
Now ∫kx=k∫x\int {kx = k\int x } , where kk is a constant. So,
∫sin⁡xcos⁡x=∫12×sin⁡2x ⇒∫sin⁡xcos⁡x=12×∫sin⁡2x...(1)   \int {\sin x\cos x} = \int {\dfrac{1}{2} \times \sin 2x} \\\ \Rightarrow \int {\sin x\cos x} = \dfrac{1}{2} \times \int {\sin 2x} ...(1) \;
We also know that –
∫sin⁡f(x)=−cos⁡f(x)f′(x)+c ⇒∫sin⁡2x=−cos⁡2xd(2x)dx+c ⇒∫sin⁡2x=−12×cos⁡2x+c  \int {\sin \\{ f(x)} \\} = \dfrac{{ - \cos \\{ f(x)\\} }}{{f'(x)}} + c \\\ \Rightarrow \int {\sin 2x = \dfrac{{ - \cos 2x}}{{\dfrac{{d(2x)}}{{dx}}}}} + c \\\ \Rightarrow \int {\sin 2x} = \dfrac{{ - 1}}{2} \times \cos 2x + c \\\
Using this value in (1), we get –
∫sin⁡xcos⁡x=12×−12×cos⁡2x+c ⇒∫sin⁡xcos⁡x=−cos⁡2x4+c   \int {\sin x\cos x} = \dfrac{1}{2} \times \dfrac{{ - 1}}{2} \times \cos 2x + c \\\ \Rightarrow \int {\sin x\cos x} = \dfrac{{ - \cos 2x}}{4} + c \;
We know, cos⁡2x=1−2sin⁡2x, cos⁡2x=2cos⁡2x−1\cos 2x = 1 - 2{\sin ^2}x,\,\cos 2x = 2{\cos ^2}x - 1
Putting these two values of cos⁡2x\cos 2x in the obtained answer, we get –
∫sin⁡xcos⁡x=−(1−2sin⁡2x)4+c ⇒∫sin⁡xcos⁡x=−14+sin⁡2x2+c and ∫sin⁡xcos⁡x=−(2cos⁡2x−1)4+c ∫sin⁡xcos⁡x=−cos⁡2x2+14+c   \int {\sin x\cos x} = \dfrac{{ - (1 - 2{{\sin }^2}x)}}{4} + c \\\ \Rightarrow \int {\sin x\cos x} = \dfrac{{ - 1}}{4} + \dfrac{{{{\sin }^2}x}}{2} + c \\\ and \\\ \int {\sin x\cos x} = \dfrac{{ - (2{{\cos }^2}x - 1)}}{4} + c \\\ \int {\sin x\cos x} = \dfrac{{ - {{\cos }^2}x}}{2} + \dfrac{1}{4} + c \;
So, the correct answer is “ −14+sin⁡2x2+c\dfrac{{ - 1}}{4} + \dfrac{{{{\sin }^2}x}}{2} + c OR −cos⁡2x2+14+c\dfrac{{ - {{\cos }^2}x}}{2} + \dfrac{1}{4} + c”.

Note : An integral that is expressed with upper and lower limits is called a definite integral while an indefinite integral is expressed without limits like in this question. The derivative of a function is unique but integral or anti-derivative of a function can be infinite, like in this question, we have found three antiderivatives of the same function. Here, varying the value of the arbitrary constant, one can get different values of integral of a function.