If (1-5 ^2x ^5x ^2x)dx=f(x)+C, where C is the constant of integration, then f(( )/(6))-f((…

Question

If $\int (\frac{1-5{\cos }^{2}x}{{\sin }^{5}x{\cos }^{2}x})dx=f(x)+C$, where C is the constant of integration, then $f(\frac{\pi }{6})-f(\frac{\pi }{4})$ is equal to

Accepted Answer

The given integral is :

$I=\int (\frac{1-5{\cos }^{2}x}{{\sin }^{5}x{\cos }^{2}x})dx$

We can rewrite the integrand by splitting the terms:

$\begin{gathered} I=\int (\frac{1}{{\sin }^{5}x{\cos }^{2}x}-\frac{5{\cos }^{2}x}{{\sin }^{5}x{\cos }^{2}x})dx \ I=\int (\frac{{\sec }^{2}x}{{\sin }^{5}x}-5{\csc }^{5}x)dx \end{gathered}$

Observe that the derivative of $f(x)=\frac{	an x}{{\sin }^{5}x}$ gives us this exact expression:

$\begin{gathered} \frac{d}{dx}(\frac{	an x}{{\sin }^{5}x})=\frac{{\sec }^{2}x\cdot {\sin }^{5}x-	an x\cdot (5{\sin }^{4}x\cos x)}{{\sin }^{10}x} \ =\frac{{\sec }^{2}x{\sin }^{5}x-5{\sin }^{5}x}{{\sin }^{10}x}=\frac{{\sec }^{2}x-5}{{\sin }^{5}x}=\frac{1-5{\cos }^{2}x}{{\sin }^{5}x{\cos }^{2}x} \end{gathered}$

Therefore, $f(x)=\frac{	an x}{{\sin }^{5}x}=\frac{\sin x/\cos x}{{\sin }^{5}x}=\frac{1}{{\sin }^{4}x\cos x}$.

Calculate $f(\frac{\pi }{6})-f(\frac{\pi }{4})$

We substitute the values into the function $f(x)=\frac{1}{{\sin }^{4}x\cos x}$ :

At $x=\frac{\pi }{6}$ :

$\begin{gathered} \sin (\frac{\pi }{6})=\frac{1}{2},\, \cos (\frac{\pi }{6})=\frac{\sqrt{3}}{2} \ f(\frac{\pi }{6})=\frac{1}{(1/2{)}^{4}\cdot (\sqrt{3}/2)}=\frac{1}{(1/16)\cdot (\sqrt{3}/2)}=\frac{32}{\sqrt{3}} \end{gathered}$

At $x=\frac{\pi }{4}$ :

$\sin (\frac{\pi }{4})=\frac{1}{\sqrt{2}},\, \cos (\frac{\pi }{4})=\frac{1}{\sqrt{2}}$

$\Rightarrow$$f(\frac{\pi }{4})=\frac{1}{(1/\sqrt{2}{)}^{4}\cdot (1/\sqrt{2})}=\frac{1}{(1/4)\cdot (1/\sqrt{2})}=4\sqrt{2}$

$	herefore $ $f(\frac{\pi }{6})-f(\frac{\pi }{4})=\frac{32}{\sqrt{3}}-4\sqrt{2}$

Factor out $\frac{4}{\sqrt{3}}$ to match the options :

$=\frac{4}{\sqrt{3}}(8-\sqrt{2}\cdot \sqrt{3})=\frac{4}{\sqrt{3}}(8-\sqrt{6})$

Solutions · 1