How do I find $\int {tan}^{4} x d x$ $inttan^4x dx$ ?

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How do I find $\int {tan}^{4} x d x$ $inttan^4x dx$ ?

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Answer 1 · 2015-03-07T23:47:13

Well, this one is quite difficult;

$\int {tan}^{4} (x) d x = \int {tan}^{2} (x) {tan}^{2} (x) d x =$ $inttan^4(x)dx=int tan^2(x)tan^2(x)dx=$
$= \int \frac{{sin}^{2} (x)}{{cos}^{2} (x)} {tan}^{2} (x) d x =$ $=intsin^2(x)/cos^2(x)tan^2(x)dx=$
$= \int \frac{1 - {cos}^{2} (x)}{{cos}^{2} (x)} {tan}^{2} (x) d x =$ $=int(1-cos^2(x))/cos^2(x)tan^2(x)dx=$
$= \int [\frac{1}{{cos}^{2} (x)} - 1] {tan}^{2} (x) d x =$ $=int[1/cos^2(x)-1]tan^2(x)dx=$
$= \int [\frac{{tan}^{2} (x)}{{cos}^{2} (x)} - {tan}^{2} (x)] d x =$ $=int[tan^2(x)/cos^2(x)-tan^2(x)]dx=$
But $d [tan (x)] = \frac{1}{{cos}^{2} (x)} d x$ $d[tan(x)]=1/cos^2(x)dx$

$= \int {tan}^{2} (x) d [tan (x)] - \int {tan}^{2} (x) d x =$ $=inttan^2(x)d[tan(x)]-inttan^2(x)dx=$
$= \frac{{tan}^{3} (x)}{3} - \int {tan}^{2} (x) d x =$ $=tan^3(x)/3-inttan^2(x)dx=$
$= \frac{{tan}^{3} (x)}{3} - \int \frac{{sin}^{2} (x)}{{cos}^{2} (x)} d x =$ $=tan^3(x)/3-intsin^2(x)/cos^2(x)dx=$
$= \frac{{tan}^{3} (x)}{3} - \int [\frac{1 - {cos}^{2} (x)}{{cos}^{2} (x)}] d x =$ $=tan^3(x)/3-int[(1-cos^2(x))/cos^2(x)]dx=$
$= \frac{{tan}^{3} (x)}{3} - {\int \frac{1}{{cos}^{2} (x)} d x - \int d x} =$ $=tan^3(x)/3-{int1/cos^2(x)dx-intdx}=$
$= \frac{{tan}^{3} (x)}{3} - tan (x) - x + c$ $=tan^3(x)/3-tan(x)-x+c$

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How do I find $\int {tan}^{4} x d x$ $inttan^4x dx$ ?

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