Question

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

• A. $\frac{\tan 7_X}{7}+c$
• B. $\frac{\sec 7x}{7}+c$
• C. $\frac{{\tan }^{7}x}{7}+c$
• D. $\frac{{\sec }^{7}x}{7}+c$

Answer 1

Answer: (C)

$\frac{{\tan }^{7}x}{7}+c$

To find $\int \frac{{\sin }^{6}x}{{\cos }^{8}x}\cdot dx=\int {\tan }^{6}x\cdot {\sec }^{2}x\cdot dx$,

substitute $\tan x=t$

${\sec }^{2}xdx=dt$

$=\int t^{6}dt=\frac{t^7}{7}+c$

$=\frac{{\tan }^{7}x}{7}+c\cdot$

$\int \frac{{\sin }^{6}x}{{\cos }^{8}x}\cdot dx=\frac{{\tan }^{7}x}{7}+c\cdot$