If _0^( )/(3) ^4xdx=a +b√3, where a and b are rational numbers, then 9a+8b is equal to :

Question

If $\int _{0}^{\frac{\pi }{3}}{\cos }^{4}xdx=a\pi +b\sqrt{3}$, where $a$ and $b$ are rational numbers, then $9a+8b$ is equal to :

Accepted Answer

To solve the given integral $\int _{0}^{\frac{\pi }{3}}{\cos }^{4}xdx$, we'll apply a known power-reduction formula that allows us to express even powers of sine and cosine functions in terms of cosine of multiple angles. Specifically for ${\cos }^{4}x$, we can write it in terms of double angles as:

${\cos }^{4}x={(\frac{1+\cos (2x)}{2})}^{2}$

We can then expand and simplify the integral using this formula. Let's proceed with this:

$\int _{0}^{\frac{\pi }{3}}{\cos }^{4}xdx=\int _{0}^{\frac{\pi }{3}}{(\frac{1+\cos (2x)}{2})}^{2}dx$

Now, let's expand the integrand and then integrate term by term:

$=\int _{0}^{\frac{\pi }{3}}(\frac{1}{4}+\frac{1}{2}\cos (2x)+\frac{1}{4}{\cos }^{2}(2x))dx$

For the ${\cos }^{2}(2x)$ term, we again use the power reduction formula:

${\cos }^{2}(2x)=\frac{1+\cos (4x)}{2}$

Let's substitute this into the integral and continue:

$=\int _{0}^{\frac{\pi }{3}}(\frac{1}{4}+\frac{1}{2}\cos (2x)+\frac{1}{4}(\frac{1+\cos (4x)}{2}))dx$

Simplify and integrate:

$=\int _{0}^{\frac{\pi }{3}}(\frac{1}{4}+\frac{1}{2}\cos (2x)+\frac{1}{8}+\frac{1}{8}\cos (4x))dx$

$=\int _{0}^{\frac{\pi }{3}}(\frac{3}{8}+\frac{1}{2}\cos (2x)+\frac{1}{8}\cos (4x))dx$

$={(\frac{3}{8}x+\frac{1}{4}\sin (2x)+\frac{1}{32}\sin (4x))|}_{\frac{\pi }{3}}^{0}$

Evaluating this from $0$ to $\frac{\pi }{3}$:

$=(\frac{3}{8}\cdot \frac{\pi }{3}+\frac{1}{4}\sin (2\cdot \frac{\pi }{3})+\frac{1}{32}\sin (4\cdot \frac{\pi }{3}))-(\frac{3}{8}\cdot 0+\frac{1}{4}\sin (0)+\frac{1}{32}\sin (0))$

$=\frac{\pi }{8}+\frac{1}{4}\sin (\frac{2\pi }{3})+\frac{1}{32}\sin (\frac{4\pi }{3})$

$\sin (\frac{2\pi }{3})$ is $\frac{\sqrt{3}}{2}$ and $\sin (\frac{4\pi }{3})$ is $-\frac{\sqrt{3}}{2}$:

$=\frac{\pi }{8}+\frac{1}{4}\cdot \frac{\sqrt{3}}{2}-\frac{1}{32}\cdot \frac{\sqrt{3}}{2}$

$=\frac{\pi }{8}+\frac{\sqrt{3}}{8}-\frac{\sqrt{3}}{64}$

Now, combining terms we get the final result:

$=\frac{\pi }{8}+\frac{8\sqrt{3}-\sqrt{3}}{64}$

$=\frac{\pi }{8}+\frac{7\sqrt{3}}{64}$

Now, let's match this result to the form $a\pi +b\sqrt{3}$ and find $a$ and $b$:

$a=\frac{1}{8},\, b=\frac{7}{64}$

Now we find $9a+8b$:

$9a+8b=9\cdot \frac{1}{8}+8\cdot \frac{7}{64}$

$=\frac{9}{8}+\frac{7}{8}$

$=\frac{16}{8}$

$=2$

Therefore, the value of $9a+8b$ is 2, which correspond to Option A.

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