Question

$\int_{}^{}\sec^{4/3}$x cosec^8/3 x dx is equal to –

• A. –$\frac{3}{5}$tan^–8/3 x + 3 tan^–2/3 x + c
• B. –$\frac{3}{5}$tan^–5/3 x + 3 tan^1/3 x + c
• C. $\frac{3}{5}$tan^–8/3 x + 3 tan^–2/3 x + c
• D. –$\frac{3}{5}$tan^–8/3 x – 3 tan^–2/3 x + c

Answer 1

Answer: (B)

–$\frac{3}{5}$tan^–5/3 x + 3 tan^1/3 x + c

Answer 2

We have, I = $\int \sec ^ { 4 / 3 }$ x cosec^8/3 x dx

= $\int \frac { 1 } { \cos ^ { 4 / 3 } x \sin ^ { 8 / 3 } x }$ dx = $\int \cos ^ { - 4 / 3 }$ x sin^–8/3 x dx

Since – $\left( \frac { 4 } { 3 } + \frac { 8 } { 3 } \right)$ = –4, which is an even integer. So, we divide both numerator and denominator by cos⁴ x.

\ I = dx

= sec² x dx = dt where t = tan x

= $\int \left( \mathrm { t } ^ { - 8 / 3 } + \mathrm { t } ^ { - 2 / 3 } \right)$ dt = – $\frac { 3 } { 5 } \mathrm { t } ^ { - 5 / 3 } + 3 \mathrm { t } ^ { 1 / 3 }$ + c

= –tan^1/3 x + c.

Hence (2) is the correct answer.