Question #61aa2

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Answer 1 · 2016-11-18T16:56:03

$\int {tan}^{3} x {sec}^{2} x d x = \frac{1}{4} {tan}^{4} x + C$ $inttan^3xsec^2xdx=1/4tan^4x+C$

$\int {tan}^{3} x {sec}^{2} x d x$ $inttan^3xsec^2xdx$

the key here is to remember that

$\frac{d}{d x} ({tan}^{n} x) = n {tan}^{n - 1} x {sec}^{2} x$ $d/(dx)(tan^nx)=ntan^(n-1)xsec^2x$

**see below**

so comparing this with the required integral

try $\frac{d}{d x} ({tan}^{4} x)$ $d/(dx)(tan^4x)$

$= 4 {tan}^{3} x {sec}^{2} x$ $=4tan^3xsec^2x$

$\int {tan}^{3} x {sec}^{2} x d x = \frac{1}{4} {tan}^{4} x + C$ $inttan^3xsec^2xdx=1/4tan^4x+C$

proof

$\frac{d}{d x} ({tan}^{n} x) = n {tan}^{n - 1} x {sec}^{2} x$ $d/(dx)(tan^nx)=ntan^(n-1)xsec^2x$

$y = {tan}^{n} x$ $y=tan^nx$

$u = tan x \Rightarrow \frac{d u}{d x} = {sec}^{2} x$ $u=tanx=>(du)/(dx)=sec^2x$

$y = u^{n} \Rightarrow \frac{d y}{d u} = n (u^{n - 1})$ $y=u^n=>(dy)/(du)=n(u^(n-1))$

$\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ $(dy)/(dx)=(dy)/(du)xx(du)/(dx)$

$\frac{d y}{d x} = n (u^{n - 1}) \times {sec}^{2} x$ $(dy)/(dx)=n(u^(n-1))xxsec^2x$

$\frac{d y}{d x} = n {tan}^{n - 1} x \times {sec}^{2} x$ $(dy)/(dx)=ntan^(n-1)x xxsec^2x$

$= n {tan}^{n - 1} x {sec}^{2} x$ $=ntan^(n-1)xsec^2x$