Question

$\int {\tan }^{6}x{\sec }^{4}xdx=$ (where $t=\tan x$)

• A. $\frac{t^9}{9}+\frac{t^7}{7}+c$
• B. $\frac{t^9}{9}-\frac{t^7}{7}+c$
• C. $\frac{t^7}{7}+\frac{t^5}{5}+c$
• D. $\frac{t^4}{4}+\frac{t^3}{3}+\frac{t^2}{2}+c$

Answer 1

The correct option is A $\frac{t^9}{9}+\frac{t^7}{7}+c$

$\int ta{n}^{6}xse{c}^{4}xdx\cdot$

$\int ta{n}^{6}x\cdot se{c}^{2}x\cdot se{c}^{2}xdx$

$t=tanx$

$dt=se{c}^{2}x\cdot dx$

$\int ta{n}^{6}x[1+ta{n}^{2}x]se{c}^{2}xdx$

$=\int t^6(1+t^2)dt$

$=\int (t^6+t^8)dt$

$=\frac{t^7}{7}+\frac{t^9}{9}+c\cdot$

where $t=tanx\cdot$

$\int ta{n}^{6}xse{c}^{4}xdx=\frac{t^7}{7}+\frac{t^9}{9}+c$ where $t=tanx$