Question

The value of
$\int{(\sin x\cdot \cos x\cdot \cos 2x\cdot \cos 4x\cdot \cos 8x\cdot \cos 16x)}dx$ is equal:

Answer 1

Hint: -Here, in this question, we will repeatedly use the half angle formula to simplify the expression and then we will integrate the simplified expression.
The most important formula that must be known to the students is as follows
The half-angle formula is as follows
$\Rightarrow \sin 2x=2\sin x\cdot \cos x$
So, for making the above expression, we will have to multiply and divide 2 with the expression that is inside the integral for using the half angle formula. We will continue the process until we can not apply the half angle formula any further.

Complete step-by-step answer:
As mentioned in the question, we have to find the value of the integration that is given in the question.
Now, as mentioned in the hint, we will first multiply and divide the expression with 2 and hence, we can write as follows

& \Rightarrow \int{(\sin x\cdot \cos x\cdot \cos 2x\cdot \cos 4x\cdot \cos 8x\cdot \cos 16x)}dx \\\ & \Rightarrow \int{\left( \dfrac{\left( 2\sin x\cdot \cos x \right)\cdot \cos 2x\cdot \cos 4x\cdot \cos 8x\cdot \cos 16x}{2} \right)}dx \\\ \end{aligned}$$ Now, applying the half angle formula, we can write as follows $$\begin{aligned} & \Rightarrow \int{\left( \dfrac{\left( 2\sin x\cdot \cos x \right)\cdot \cos 2x\cdot \cos 4x\cdot \cos 8x\cdot \cos 16x}{2} \right)}dx \\\ & \left( 2\sin x\cdot \cos x=\sin 2x \right) \\\ & \Rightarrow \int{\left( \dfrac{\sin 2x\cdot \cos 2x\cdot \cos 4x\cdot \cos 8x\cdot \cos 16x}{2} \right)}dx \\\ \end{aligned}$$ Now, again we will multiply and divide the expression with 2 and then we will apply the half angle formula as follows $$\begin{aligned} & \Rightarrow \int{\left( \dfrac{2\sin 2x\cdot \cos 2x\cdot \cos 4x\cdot \cos 8x\cdot \cos 16x}{4} \right)}dx \\\ & \left( 2\sin 2x\cdot \cos 2x=\sin 4x \right) \\\ & \Rightarrow \int{\left( \dfrac{\sin 4x\cdot \cos 4x\cdot \cos 8x\cdot \cos 16x}{4} \right)}dx \\\ \end{aligned}$$ Now, again we will multiply and divide the expression with 2 and then we will apply the half angle formula as follows $$\begin{aligned} & \Rightarrow \int{\left( \dfrac{2\sin 4x\cdot \cos 4x\cdot \cos 8x\cdot \cos 16x}{8} \right)}dx \\\ & \left( 2\sin 4x\cdot \cos 4x=\sin 8x \right) \\\ & \Rightarrow \int{\left( \dfrac{\sin 8x\cdot \cos 8x\cdot \cos 16x}{8} \right)}dx \\\ \end{aligned}$$ Now, again we will multiply and divide the expression with 2 and then we will apply the half angle formula as follows $$\begin{aligned} & \Rightarrow \int{\left( \dfrac{2\sin 8x\cdot \cos 8x\cdot \cos 16x}{16} \right)}dx \\\ & \left( 2\sin 8x\cdot \cos 8x=\sin 16x \right) \\\ & \Rightarrow \int{\left( \dfrac{\sin 16x\cdot \cos 16x}{16} \right)}dx \\\ \end{aligned}$$ Now, for the final time, again we will multiply and divide the expression with 2 and then we will apply the half angle formula as follows $$\begin{aligned} & \Rightarrow \int{\left( \dfrac{2\sin 16x\cdot \cos 16x}{32} \right)}dx \\\ & \left( 2\sin 16x\cdot \cos 16x=\sin 32x \right) \\\ & \Rightarrow \int{\left( \dfrac{\sin 32x}{32} \right)}dx \\\ \end{aligned}$$ Now, as we have simplified the expression, we can easily integrate the above expression as follows $$\begin{aligned} & \Rightarrow \int{\left( \dfrac{\sin 32x}{32} \right)}dx \\\ & \Rightarrow \dfrac{1}{32}\left[ \dfrac{-\cos 32x}{32} \right]+C \\\ & \Rightarrow \left[ \dfrac{-\cos 32x}{{{\left( 32 \right)}^{2}}} \right]+C \\\ \end{aligned}$$ (Because we know that integral of $$\sin nx$$ is $$\dfrac{-\cos \left( nx \right)}{n}$$ and C is the constant of integration) Note:-For this question, it is very important to know the half angle formula that is as follows $$\Rightarrow \sin 2x=2\sin x\cdot \cos x$$ As without knowing it, one can never get to the correct answer as the simplification of given expression is not at all easy otherwise. Also, it is essential to know that the integral of $$\sin nx$$ is $$\dfrac{-\cos \left( nx \right)}{n}$$ . Another thing that is very important is that the students must put the constant of integration at the end of the value.