How do you find the integral of $\int sin x \cdot tan x d x$ $int sin x * tan x dx$ ?

Question

63555 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-08T08:08:23

The answer is $= ln (| tan x + sec x |) - sin x + C$ $=ln(|tanx+secx|)-sinx+C$

We need

$tan x = \frac{sin x}{cos x}$ $tanx=sinx/cosx$

$\int sec x d x = ln (tan x + sec x) + C$ $intsecxdx=ln(tanx+secx)+C$

Therefore,

$\int sin x tan x d x = \int sec x {sin}^{2} x d x = \int sec x (1 - {cos}^{2} x) d x$ $intsinxtanxdx=intsecxsin^2xdx=intsecx(1-cos^2x)dx$

$= \int (sec x - cos x) d x$ $=int(secx-cosx)dx$

$= \int sec x d x - \int cos x d x$ $=intsecxdx-intcosxdx$

$= ln (| tan x + sec x |) - sin x + C$ $=ln(|tanx+secx|)-sinx+C$

How do you find the integral of int sin x * tan x dx∫sinx⋅tanxdxint sin x * tan x dx?