Question

$\int \frac{si{n}^{4}x}{co{s}^{8}x}dx$ is equal to (where $C$ is an arbitrary constant)

• A. $\frac{\left(1+{\left(tan\right)}^{5}x\right)}{5}+\frac{{\left(tan\right)}^{5}x}{7}+C$
• B. $\frac{ta{n}^{5}x}{5}+\frac{ta{n}^{7}x}{7}+C$
• C. $\frac{ta{n}^{7}x}{5}+\frac{ta{n}^{5}x}{7}+C$
• D. None of these

Answer 1

Answer: (B)

$\int \frac{\frac{si{n}^{4}x}{co{s}^{4}x}}{\frac{co{s}^{8}x}{co{s}^{4}x}}.dx=\int ta{n}^{4}x.se{c}^{4}xdx$
$=\int {\left(tan\right)}^{4}x\left(1+{\left(tan\right)}^{2}x\right){\left(sec\right)}^{2}xdx$
$=\int ta{n}^{4}x\left(1+ta{n}^{2}x\right)d(tanx)=\int t^4{\left(1+t^2\right)}^{}dt=\frac{t^5}{5}+\frac{t^7}{7}+\text{C}$