Question

$\displaystyle \int \frac{sin^{4} x}{cos^{8} ⁡ x} \, dx$ 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. $\frac{tan^{5} x}{5}+\frac{tan^{7} ⁡ x}{7}+C$
• C. $\frac{tan^{7} x}{5}+\frac{tan^{5} ⁡ x}{7}+C$
• D. None of these

Answer 1

Answer: (B)

$\displaystyle \int \frac{\frac{sin^{4} x}{cos^{4} ⁡ x}}{\frac{cos^{8} ⁡ x}{cos^{4} ⁡ x}}.dx=\displaystyle \int tan^{4} ⁡ x.sec^{4} ⁡ x \, dx$
$=\displaystyle \int \left(tan\right)^{4} x\left(1 + \left(tan\right)^{2} ⁡ x\right)\left(sec\right)^{2} ⁡ x \, dx$
$=\displaystyle \int tan^{4} x \left(1 + tan^{2} ⁡ x\right) d (tan x) = \displaystyle \int t^{4} \left(1 + t^{2}\right)^{ } dt= \frac{t^{5}}{5} + \frac{t^{7}}{7} + \text{C}$