Question

If the value of integral $\int \frac{{\sin }^{6}x}{{\cos }^{8}x}dxisk{\tan }^{7}x+c,$ then k = _________________.

Answer 1

$I=\int \frac{{\sin }^{6}x}{{\cos }^{8}x}dx$

$\Rightarrow I=\int \frac{{\sin }^{6}x}{{\cos }^{6}x\times {\cos }^{2}x}dx$

$\Rightarrow I=\int {\tan }^{6}x{\sec }^{2}xdx$

Let $\tan x=z$

$\Rightarrow {\sec }^{2}xdx=dz$

$\therefore I=\int z^{6}dz$

$\Rightarrow I=\frac{z^7}{7}+C$

$\Rightarrow I=\frac{1}{7}{\tan }^{7}x+C$

Comparing with the given expression, we get

k = $\frac{1}{7}$


If the value of integral $\int \frac{{\sin }^{6}x}{{\cos }^{8}x}dx\mathrm{is}k{\tan }^{7}x+c,$ then k = $\underline{\frac{1}{7}}$.