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Question: How do you integrate \[\int {x\sin x\cos x} \] by integration by parts method?...

How do you integrate ∫xsin⁡xcos⁡x\int {x\sin x\cos x} by integration by parts method?

Explanation

Solution

Hint : The above question is based on the concept of integration. Since it is an indefinite integral which has no upper and lower limits, we can apply integration properties by integrating it in parts so that we can find the antiderivative of the above expression and also by applying multiple angle formulas.

Complete step-by-step answer :
Integration is a way of finding the antiderivative of any function. It is the inverse of differentiation. It denotes the summation of discrete data. The calculation of such small problems is an easy task but for adding big problems which include higher limits, integration method is used. The above given expression is an indefinite integral which means there are no upper or lower limits given.
The above equation should be in the form given below.
∫f(x)=F(x)+C\int {f\left( x \right) = F(x) + C}
where C is constant.
Now the given expression contains trigonometric functions. So, we need to bring the expression into multiple angle trigonometric functions. This can be done by the following way
∫xsin⁡xcos⁡xdx=12∫xsin⁡2xdx\int {x\sin x\cos xdx = \dfrac{1}{2}\int {x\sin 2xdx} }
Further we need to integrate the integral by parts
∫uv′=uv−∫u′v\int {uv'} = uv - \int {u'v}
where u=xu = x
⇒u′=1\Rightarrow u' = 1
Now v′=sin⁡2xv' = \sin 2x
∴v=−cos⁡2x2\therefore v= \dfrac{{ - \cos 2x}}{2}
Now by further solving,

∫xsin⁡xcos⁡xdx=−xcos⁡x2+12∫cos⁡2xdx =−xcos⁡2x2+12×sin⁡2x2 =sin⁡2x4−xcos⁡2x2   \int {x\sin x\cos xdx = - \dfrac{{x\cos x}}{2} + \dfrac{1}{2}\int {\cos 2xdx} } \\\ = - \dfrac{{x\cos 2x}}{2} + \dfrac{1}{2} \times \dfrac{{\sin 2x}}{2} \\\ = \dfrac{{\sin 2x}}{4} - \dfrac{{x\cos 2x}}{2} \;

Then the final step we get is

∫xsin⁡xcos⁡xdx=12(sin⁡2x4−xcos⁡2x2)+C =sin⁡2x8−xsin⁡2x4+C   \int {x\sin x\cos xdx = \dfrac{1}{2}\left( {\dfrac{{\sin 2x}}{4} - \dfrac{{x\cos 2x}}{2}} \right) + C} \\\ = \dfrac{{\sin 2x}}{8} - \dfrac{{x\sin 2x}}{4} + C \;

Therefore, we get the above solution.
So, the correct answer is “sin⁡2x8−xsin⁡2x4+C\dfrac{{\sin 2x}}{8} - \dfrac{{x\sin 2x}}{4} + C”.

Note : An important thing to note is that the double angle trigonometric function of sine function is
sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x .So in the expression we multiply 2 in the numerator and denominator so that we get the above formula for double angle and substitute that term in the numerator with sin2x so that it becomes easy to integrate.