$\int \frac{(4 + 5 tan (x)) {sec}^{4} (x)}{{tan}^{5} (x)} d x$ $int((4+5 tan (x)) sec^4 (x)) / tan^5(x)dx$ ?

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$\int \frac{(4 + 5 tan (x)) {sec}^{4} (x)}{{tan}^{5} (x)} d x$ $int((4+5 tan (x)) sec^4 (x)) / tan^5(x)dx$ ?

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Answer 1 · 2018-03-03T16:54:15

$\int \frac{(4 + 5 tan x) {sec}^{4} x}{{tan}^{5} x} d x = - \frac{3 + 5 tan x + 6 {tan}^{2} x + 15 {tan}^{3} x}{3 {tan}^{4} x} + C$ $int ((4+5tanx)sec^4x)/tan^5x dx = -(3+5tanx+6tan^2x+15tan^3x)/(3tan^4x)+C$

Explanation:

Using the trigonometric identity:

$1 + {tan}^{2} α = 1 + \frac{{sin}^{2} α}{{cos}^{2} α} = \frac{{cos}^{2} α + {sin}^{2} α}{{cos}^{2} α} = {sec}^{2} α$ $1+tan^2alpha = 1+sin^2alpha/cos^2alpha = (cos^2alpha+sin^2alpha)/cos^2alpha = sec^2alpha$

At the numerator let:

${sec}^{4} x = {sec}^{2} x {sec}^{2} x = (1 + {tan}^{2} x) {sec}^{2} x$ $sec^4x = sec^2x sec^2x = (1+tan^2x)sec^2x$

Then substitute $t = tan x$ $t= tanx$ , $d t = {sec}^{2} x d x$ $dt = sec^2xdx$ :

$\int \frac{(4 + 5 tan x) {sec}^{4} x}{{tan}^{5} x} d x = \int \frac{(4 + 5 t) (1 + t^{2})}{t^{5}} d t$ $int ((4+5tanx)sec^4x)/tan^5x dx = int ((4+5t)(1+t^2))/t^5dt$

$\int \frac{(4 + 5 tan x) {sec}^{4} x}{{tan}^{5} x} d x = \int \frac{4 + 5 t + 4 t^{2} + 5 t^{3}}{t^{5}} d t$ $int ((4+5tanx)sec^4x)/tan^5x dx = int (4+5t+4t^2+5t^3)/t^5dt$

and using the linearity of the integral:

$\int \frac{(4 + 5 tan x) {sec}^{4} x}{{tan}^{5} x} d x = 4 \int \frac{d t}{t^{5}} + 5 \int \frac{d t}{t^{4}} + 4 \int \frac{d t}{t^{3}} + 5 \int \frac{d t}{t^{2}}$ $int ((4+5tanx)sec^4x)/tan^5x dx = 4 int (dt)/t^5 +5 int (dt)/t^4+4int (dt)/t^3+5int (dt)/t^2$

$\int \frac{(4 + 5 tan x) {sec}^{4} x}{{tan}^{5} x} d x = - \frac{1}{t^{4}} - \frac{5}{3 t^{3}} - \frac{2}{t^{2}} - \frac{5}{t} + C$ $int ((4+5tanx)sec^4x)/tan^5x dx = -1/t^4-5/(3t^3)-2/t^2-5/t+C$

$\int \frac{(4 + 5 tan x) {sec}^{4} x}{{tan}^{5} x} d x = - \frac{3 + 5 t + 6 t^{2} + 15 t^{3}}{3 t^{4}} + C$ $int ((4+5tanx)sec^4x)/tan^5x dx = -(3+5t+6t^2+15t^3)/(3t^4)+C$

and undoing the substitution:

$\int \frac{(4 + 5 tan x) {sec}^{4} x}{{tan}^{5} x} d x = - \frac{3 + 5 tan x + 6 {tan}^{2} x + 15 {tan}^{3} x}{3 {tan}^{4} x} + C$ $int ((4+5tanx)sec^4x)/tan^5x dx = -(3+5tanx+6tan^2x+15tan^3x)/(3tan^4x)+C$

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$\int \frac{(4 + 5 tan (x)) {sec}^{4} (x)}{{tan}^{5} (x)} d x$ $int((4+5 tan (x)) sec^4 (x)) / tan^5(x)dx$ ?

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Explanation:

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∫(4+5tan(x))sec4(x)tan5(x)dxint((4+5 tan (x)) sec^4 (x)) / tan^5(x)dx?

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Explanation:

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$\int \frac{(4 + 5 tan (x)) {sec}^{4} (x)}{{tan}^{5} (x)} d x$ $int((4+5 tan (x)) sec^4 (x)) / tan^5(x)dx$ ?