Integration of Trigonometric Functions
Integration of Trigonometric Functions
Example 1:
$$\int \tan x \, dx $$
Solution:
\begin{align*}
\int \tan x \, dx &= \int \frac{\sin x }{\cos x }\, dx\\\\
\textup{Let}\,\, u&=\cos x\\
du &= -\sin x dx\\
-du &= \sin x dx\\\\
\int \tan x\, dx &= \int \frac{-du}{u}\\
&=- \int \frac{1}{u}\,du\\
&= -\ln \lvert u \lvert+C\\
&= -\ln \lvert \cos x \lvert+C\\
&= \ln 1 -\ln \lvert \cos x \lvert+C\\
&= \ln \frac{1}{\lvert \cos x \lvert}+C\\
&= \ln \lvert \sec x \lvert+C\\
\end{align*}
Example 2:
$$\int \sin^4 x \cos x\, dx $$
Solution:
- $$\int \sin^4 x \cos x\, dx = \int (\sin x )^4\cos x \, xdx$$
- $$\textup{Let}\,\, u=\sin x$$
- $$du = \cos x dx$$
- $$\int \sin^4 x \cos x\, dx = \int u^4 \,du $$
- $$= \frac{u^{4+1}}{4+1}+C$$
- $$= \frac{u^{5}}{5}+C$$
- $$= \frac{1}{5} \, u^5+C$$
- $$= \frac{1}{5} \, (\sin x)^5+C$$
- $$= \frac{1}{5} \, \sin^5 x+C$$