Question

Evaluate: $\int sec^{4/3} \,x cosec^{8/3}\,x \,dx$

• A. $\frac{3}{5} tan^{-5/3} x - 3 tan^{1/3} x + C$
• B. $- \frac{3}{5} tan^{-5/3} x - 3 tan^{1/3} x + C$
• C. $- \frac{3}{5} tan^{-5/3} x - 3 tan^{1/3} x + C$
• D. None of these

Answer 1

Answer: (B)

$- \frac{3}{5} tan^{-5/3} x - 3 tan^{1/3} x + C$

Answer 2

Let $I= \int sec^{4/3} x \,cosec^{8/3} x \,dx$ $\Rightarrow I= \int cos^{-4/3} x \,sin^{-8/3} x \,dx$ $\Rightarrow I= \int \frac{sec^{4} \,x}{tan^{\frac{8}{3}} \,x}dx = \int \frac{\left(1+tan^{2} x\right)}{tan^{\frac{8}{3}} x} sec^{2} \,x\, dx$ Putting $tanx = t \Rightarrow sec^{2}\,x\, dx = dt$ $I = \int \frac{1+t^{2}}{t^{\frac{8}{3}}} dt = \int \left(t^{-\frac{8}{3}} + t^{-\frac{2}{3}}\right) dt$ $\Rightarrow I=\frac{-3}{5}t^{-\frac{5}{3}} + 3t^{\frac{1}{3}} + C$ $= \frac{-3}{5} tan^{-5/3} x + 3 tan^{1/3} x + C$