#int((4+5 tan (x)) sec^4 (x)) / tan^5(x)dx#?

Question

Evelyn Agard · Accepted Answer

#int ((4+5tanx)sec^4x)/tan^5x dx = -(3+5tanx+6tan^2x+15tan^3x)/(3tan^4x)+C#

Using the trigonometric identity:

#1+tan^2alpha = 1+sin^2alpha/cos^2alpha = (cos^2alpha+sin^2alpha)/cos^2alpha = sec^2alpha#

At the numerator let:

#sec^4x = sec^2x sec^2x = (1+tan^2x)sec^2x#

Then substitute #t= tanx#, #dt = sec^2xdx#:

#int ((4+5tanx)sec^4x)/tan^5x dx = int ((4+5t)(1+t^2))/t^5dt#

#int ((4+5tanx)sec^4x)/tan^5x dx = int (4+5t+4t^2+5t^3)/t^5dt#

and using the linearity of the integral:

#int ((4+5tanx)sec^4x)/tan^5x dx = 4 int (dt)/t^5 +5 int (dt)/t^4+4int (dt)/t^3+5int (dt)/t^2#

#int ((4+5tanx)sec^4x)/tan^5x dx = -1/t^4-5/(3t^3)-2/t^2-5/t+C#

#int ((4+5tanx)sec^4x)/tan^5x dx = -(3+5t+6t^2+15t^3)/(3t^4)+C#

and undoing the substitution:

#int ((4+5tanx)sec^4x)/tan^5x dx = -(3+5tanx+6tan^2x+15tan^3x)/(3tan^4x)+C#

Elijah Abram · Answer

undefined To solve the integral $\int \frac{{4 + 5 \tan(x)}}{{\tan^5(x)}} \sec^4(x) \, dx$, we can use the substitution method. Let $u = \tan(x)$. Then, $du = \sec^2(x) \, dx$ and $\sec^2(x) = 1 + \tan^2(x) = 1 + u^2$.

So the integral becomes:

$$
\int \frac{{4 + 5u}}{{u^5}} (1 + u^2) \, du
$$

Expanding and separating the terms, we get:

$$
\int \left(\frac{4}{{u^5}} + \frac{5u}{{u^5}} + \frac{4u^2}{{u^5}} + \frac{5u^3}{{u^5}}\right) \, du
$$

Now integrate each term separately:

$$
\int \frac{4}{{u^5}} \, du = -\frac{4}{{4u^4}} = -\frac{1}{{u^4}}
$$

$$
\int \frac{5u}{{u^5}} \, du = \frac{5}{{4u^4}}
$$

$$
\int \frac{4u^2}{{u^5}} \, du = \frac{4}{{-3u^3}} = -\frac{4}{{3u^3}}
$$

$$
\int \frac{5u^3}{{u^5}} \, du = \frac{5}{{-2u^2}} = -\frac{5}{{2u^2}}
$$

Now, combine the results:

$$
-\frac{1}{{u^4}} + \frac{5}{{4u^4}} - \frac{4}{{3u^3}} - \frac{5}{{2u^2}} + C
$$

Finally, replace $u$ with $\tan(x)$:

$$
-\frac{1}{{\tan^4(x)}} + \frac{5}{{4\tan^4(x)}} - \frac{4}{{3\tan^3(x)}} - \frac{5}{{2\tan^2(x)}} + C
$$