How do I intrgate the following when $u = sinx$
$\int_{}^{} \frac{1+\sin(x)}{{\cos(x)}} \ dx$
I have made a start:
$du = \cos(x)$
Therefore $\frac{dx}{du}$ = $\frac{1}{\cos(x)}$
$\int_{}^{} \frac{1}{{\cos(x)}} dx$ + $\int_{}^{} \frac{\sin(x)}{{\cos(x)}} \ dx$
How would I finish this off?
On
Substitute $$t=1+\sin(x)$$ then we get $$dt=\cos(x)dx$$ so $$dx=\frac{dt}{\cos(x)}$$ and out integral will be $$\int\frac{tdt}{1-(t-1)^2}$$
On
No substitution needed.
\begin{align*} \int \frac{1}{\cos x} dx &= \int \sec x dx \\ &=\int \frac{\sec x (\sec x +\tan x)}{\sec x +\tan x} dx \\ &=\log |\sec x + \tan x|,\end{align*} and \begin{align*} \int \frac{\sin x}{\cos x} dx &=-\log|\cos x| \\ &=\log|\sec x|. \end{align*} Add that constant of integration and call it good.
On
If you aren't required to use a substitution and you know the antiderivatives table, you may proceed as follows :
$$\int \frac{1+\sin x}{\cos x} dx =\int sec x dx+\int tan x dx=\ln |\tan x + \sec x|-\ln |cos x| +C $$
EDIT: Sniped by 15 seconds.
On
Put $x = 2t \implies 1 +\sin x = 1+\sin (2t)=(\sin t + \cos t)^2$, $\cos x = \cos (2t)=(\cos t + \sin t)(\cos t - \sin t)\implies \dfrac{1+\sin x}{\cos x} = \dfrac{\sin t +\cos t}{\cos t - \sin t}\implies I = \displaystyle -2\int \dfrac{du}{u}, u = \cos t - \sin t\implies I = -2\ln|u|+C = -2\ln|\cos t - \sin t| + C=-2\ln|\cos (x/2) - \sin (x/2)|+C$ .
On
Another method is
\begin{align}\frac{1+\sin x}{{\cos x}}&=\left(\frac{1+\sin x}{{\cos x}}\right)\frac{1-\sin x}{1-\sin x}\\&=\frac{1-\sin^2 x}{\cos x(1-\sin x)}\\&=\frac{\cos^2 x}{\cos x(1-\sin x)}\\&=\frac{\cos x}{1-\sin x}\end{align}
then you can directly integrate
$$\int \frac{1+\sin x}{{\cos x}} \ dx=\int \frac{\cos x}{1-\sin x}\,dx=-\ln|1-\sin x|+C$$
by the susbtitution $u=1-\sin x$ which implies that $du=-\cos x dx$.
It might be useful to transform the integrand first
$$\frac{1+\sin x}{\cos x} =\frac{1-\sin^2 x}{\cos x(1-\sin x)} = \frac{\cos x}{1-\sin x}$$
N0w, substitute $t= \sin x$.