Integration Techniques for Various Trigonometric Functions
Overview
When solving integration problems, you often have to integrate trigonometric functions. And, getting familiar with these integration methods, trigonometric functions can be integrated as quickly as polynomial functions.
Integration of Secant Functions, Integration of Cosecant Functions
$$ \begin{align*} \int \sec x dx =& \int \frac { \sec x (\sec x +\tan x ) }{ (\sec x +\tan x ) }dx \\ =& \int \frac { \sec^{ 2 }x+\sec x \tan x }{ \tan x +\sec x }dx \end{align*} $$ Since $ (\tan x )\prime =\sec^{ 2 }x$ and $(\sec x )\prime =\sec x \tan x$, $$ \int \sec x dx=\ln|\tan x +\sec x |+C $$
By the same method, the following is obtained. $$ \int \csc x dx=-\ln|\cot x+\csc x |+C $$
Integration of Tangent Functions, Integration of Cotangent Functions
$$ \begin{align*} \int \tan x dx =& \int \frac { \sin x }{ \cos x }dx \\ =& \int \frac { -(-\sin x ) }{ \cos x }dx \end{align*} $$
Because of $(\cos x )\prime =-\sin x$,
$$ \int \tan x dx=-\ln|\cos x |+C $$
By the same method, the following is obtained.
$$ \int \cot xdx=\ln|\sin x |+C $$
Integration of Square of Sine Functions, Integration of Square of Cosine Functions
$$ \begin{align*} \int \sin^{ 2 }xdx =& \int \frac { 1-\cos2x }{ 2 }dx \\ =& \frac { 1 }{ 2 }(x-\frac { 1 }{ 2 }\sin2x)+C \\ =& \frac { 1 }{ 4 }(2x-\sin2x)+C \end{align*} $$
By the same method, the following is obtained.
$$ \int \cos^{ 2 }xdx=\frac { 1 }{ 4 }(2x+\sin2x)+C $$
If the order is higher than $3$, reduce the order using $\sin^2 \theta + \cos^2 \theta =1$.
Integration of the product of Sine Functions and $x$, Integration of the product of Cosine Functions and $x$
$$ \int x\sin x dx = -x\cos x -\int -\cos x dx=\sin x -x\cos x +C $$
$$ \int x\cos x dx=x\sin x -\int \sin x dx=\cos x +x\sin x +C $$
Integration of the n-th power of Sine Functions and the product of Cosine Functions, Integration of the n-th power of Cosine Functions and the product of Sine Functions
When substituting $\int \sin^{ n }x\cos x dx$ for $\sin x =t$,
$$ \int t^{ n }dt=\displaystyle \frac { \sin^{ n+1 }x }{ n+1 }+C $$
When substituting $\int \cos^{ n }x\sin x dx$ for $\cos x =t$,
$$ -\int t^{ n }dt=\displaystyle -\frac { \cos^{ n+1 }x }{ n+1 }+C $$