Question

What is the integral value of $\displaystyle{\int\limits^\pi_0 {(1-|\sin8x|} \, dx }$ ?

Answer 1

$\displaystyle\int\limits_0^{\pi}(1-|\sin(8x)|)\ dx=\int\limits_0^{\pi}1\ dx-\int\limits_0^{\pi}|\sin (8x)|\ dx$

We know that,

$\displaystyle\int\limits_0^{\pi}1\ dx=\Bigg[x\Bigg]_0^{\pi}=\pi$

But what about $\displaystyle\int\limits_0^{\pi}|\sin (8x)|\ dx\ ?$

Here I split $\displaystyle\int\limits_0^{\pi}|\sin (8x)|\ dx$ into many integrals as follows:

$\displaystyle\int\limits_0^{\pi}|\sin (8x)|\ dx=\int\limits_0^{\frac{\pi}{8}}\sin (8x)\ dx-\int\limits_{\frac {\pi}{8}}^{\frac{\pi}{4}}\sin (8x)\ dx+\int\limits_{\frac {\pi}{4}}^{\frac{3\pi}{8}}\sin (8x)\ dx-\int\limits_{\frac {3\pi}{8}}^{\frac{\pi}{2}}\sin (8x)\ dx+\int\limits_{\frac {\pi}{2}}^{\frac{5\pi}{8}}\sin (8x)\ dx-\int\limits_{\frac {5\pi}{8}}^{\frac{3\pi}{4}}\sin (8x)\ dx+\int\limits_{\frac {3\pi}{4}}^{\frac{7\pi}{8}}\sin (8x)\ dx-\int\limits_{\frac {7\pi}{8}}^{\pi}\sin (8x)\ dx$

It seems much lengthy, but each definite integral values the same!

Thus the RHS can be reduced as,

$\displaystyle\int\limits_0^{\pi}|\sin (8x)|\ dx=8\int\limits_0^{\frac{\pi}{8}}\sin (8x)\ dx$

Consider the indefinite integral,

$\displaystyle\int\sin (8x)\ dx$

Let,

$u=8x\implies\dfrac {du}{dx}=8\implies dx=\dfrac{1}{8}du$

Then,

$\displaystyle\int\sin (8x)\ dx=\dfrac {1}{8}\int\sin u\ du\\\\\\\int\sin (8x)\ dx=-\dfrac {\cos (8x)}{8}$

Thus,

$\displaystyle\int\limits_0^{\pi}|\sin (8x)|\ dx=-\Bigg[\cos (8x)\Bigg]_0^{\frac{\pi}{8}}\\\\\\\int\limits_0^{\pi}|\sin (8x)|\ dx=-[\cos (\pi)-\cos (0)]\\\\\\\int\limits_0^{\pi}|\sin (8x)|\ dx=2$

Hence, finally,

$\large\underline{\underline {\displaystyle\int\limits_0^{\pi}(1-|\sin(8x)|)\ dx=\boxed {\pi-2}}}$

Answer 2

Answer:

Answer is in attachment:-

Step-by-step explanation:

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