# Question #8c487

Mar 1, 2016

$\setminus \int \setminus {\tan}^{5} \left(7 x\right) \setminus {\sec}^{4} \left(7 x\right) \mathrm{dx} = \setminus \frac{1}{7} \left(\setminus \frac{1}{8} \setminus {\sec}^{8} \left(7 x\right) - \setminus \frac{1}{3} \setminus {\sec}^{6} \left(7 x\right) + \setminus \frac{1}{4} \setminus {\sec}^{4} \left(7 x\right)\right) + C$

#### Explanation:

$\setminus \int \setminus {\tan}^{5} \left(7 x\right) . {\sec}^{4} \left(7 x\right) \mathrm{dx}$

We know,
$\setminus {\tan}^{5} \left(7 x\right) = \setminus {\tan}^{4} \left(7 x\right) . \tan \left(7 x\right)$ i.e. applying the algebric property ${x}^{a} = {x}^{a - 1} . x$

$= \setminus \int \setminus {\tan}^{4} \left(7 x\right) \setminus \tan \left(7 x\right) \setminus {\sec}^{4} \left(7 x\right) \mathrm{dx}$

Also,
$\setminus {\tan}^{4} \left(7 x\right) = {\left(\setminus {\tan}^{2} \left(7 x\right)\right)}^{2}$

$= \setminus \int {\left(\setminus {\tan}^{2} \left(7 x\right)\right)}^{2} \setminus \tan \left(7 x\right) \setminus {\sec}^{4} \left(7 x\right) \mathrm{dx}$

Now,using the identity : $\setminus {\tan}^{2} \left(x\right) = - 1 + \setminus {\sec}^{2} \left(x\right)$

$= \setminus \int {\left(- 1 + \setminus {\sec}^{2} \left(7 x\right)\right)}^{2} \setminus \tan \left(7 x\right) \setminus {\sec}^{4} \left(7 x\right) \mathrm{dx}$

Applying integral substitution as: $\setminus \int f \left(g \left(x\right)\right) \setminus \cdot {g}^{'} \left(x\right) \mathrm{dx} = \setminus \int f \left(u\right) \mathrm{du} , \quad u = g \left(x\right)$
$\setminus \sec \left(7 x\right) = u \setminus \quad \mathrm{dx} = \setminus \frac{1}{7 \setminus \tan \left(7 x\right) \setminus \sec \left(7 x\right)} \mathrm{du}$ as shown below,

Substituting $\sec \left(7 x\right) = u$
$= \setminus \int {\left(- 1 + {u}^{2}\right)}^{2} \setminus \tan \left(7 x\right) {u}^{4} \setminus \frac{1}{7 \setminus \tan \left(7 x\right) u} \mathrm{du}$

$= \setminus \int \setminus \frac{1}{7} {u}^{3} {\left({u}^{2} - 1\right)}^{2} \mathrm{du}$

Taking the constant out as: $\setminus \int a \setminus \cdot f \left(x\right) \mathrm{dx} = a \setminus \cdot \setminus \int f \left(x\right) \mathrm{dx}$

$= \setminus \frac{1}{7} \setminus \int {u}^{3} {\left({u}^{2} - 1\right)}^{2} \mathrm{du}$

Expanding ${u}^{3} {\left({u}^{2} - 1\right)}^{2}$

$= \setminus \frac{1}{7} \setminus \int \left({u}^{7} - 2 {u}^{5} + {u}^{3}\right) \mathrm{du}$

Applying sum rule as: $\setminus \int f \left(x\right) \setminus \pm g \left(x\right) \mathrm{dx} = \setminus \int f \left(x\right) \mathrm{dx} \setminus \pm \setminus \int g \left(x\right) \mathrm{dx}$

$= \setminus \frac{1}{7} \left(\setminus \int {u}^{7} \mathrm{du} - \setminus \int 2 {u}^{5} \mathrm{du} + \setminus \int {u}^{3} \mathrm{du}\right)$

Now,we know
$\setminus \int {u}^{7} \mathrm{du} = \setminus \frac{{u}^{8}}{8}$
$\setminus \int 2 {u}^{5} \mathrm{du} = \setminus \frac{{u}^{6}}{3}$ ($2 \frac{{u}^{6}}{6}$ =${u}^{6} / 3$)
$\setminus \int {u}^{3} \mathrm{du} = \setminus \frac{{u}^{4}}{4}$

i.e
$= \setminus \frac{1}{7} \left(\setminus \frac{{u}^{8}}{8} - \setminus \frac{{u}^{6}}{3} + \setminus \frac{{u}^{4}}{4}\right)$

Substituting back $u = \sec \left(7 x\right)$,
$= \setminus \frac{1}{7} \left(\setminus \frac{\setminus {\sec}^{8} \left(7 x\right)}{8} - \setminus \frac{\setminus {\sec}^{6} \left(7 x\right)}{3} + \setminus \frac{\setminus {\sec}^{4} \left(7 x\right)}{4}\right)$

Simplifying it,
$= \setminus \frac{1}{7} \left(\setminus \frac{1}{8} \setminus {\sec}^{8} \left(7 x\right) - \setminus \frac{1}{3} \setminus {\sec}^{6} \setminus \le f t \left(7 x\right) + \setminus \frac{1}{4} \setminus {\sec}^{4} \left(7 x\right)\right)$

Finally adding constant to the solution,

$= \setminus \frac{1}{7} \left(\setminus \frac{1}{8} \setminus {\sec}^{8} \left(7 x\right) - \setminus \frac{1}{3} \setminus {\sec}^{6} \setminus \le f t \left(7 x\right) + \setminus \frac{1}{4} \setminus {\sec}^{4} \left(7 x\right)\right)$ + C