Need help with integration...

Question

I need some help with this problem:

$$\int\frac 1 {1+\sin x} \, dx$$

I know it seems quite basic, but I cannot really find the right substitution to use.

Thank you in advance!

Answer 1

Hint: $$\int\frac 1 {1+\sin x} \, dx=\int\frac {1-\sin x} {\cos^2x} \, dx$$

Answer 2

$$\int\frac 1 {1+\sin x} \, dx=\int\frac {\sec^2\left(\frac x2\right)} {\left(\tan\left(\frac x2\right)+1\right)^2} \, dx$$

Here, you can use $u$-substitution where $u=\tan\left(\dfrac x2\right)+1$ and $dx=\dfrac2{\sec^2\left(\frac x2\right)du}$

$$2\int \dfrac 1{u^2}=-\dfrac{2}u$$ Plugging in $u=\tan\left(\dfrac x2\right)+1$, you get $-\dfrac{2}{\tan\left(\frac x2\right)+1}$. $$\int\frac 1 {1+\sin x}=-\dfrac2{\tan\left(\frac x2\right)+1}+C$$

Answer 3

$$\int\frac 1 {1+\sin x} \, dx =\int\frac {1-\sin x} {(1+\sin x)(1-\sin x)} \, dx=$$

$$\int\frac {1-\sin x} {\cos^2 x} dx=$$

$$\int\frac {1} {\cos^2 x} dx -\int\frac {\sin x} {\cos^2 x}=$$

$$\int \sec^2 x dx -\int\frac {\sin x} {\cos^2 x} dx$$

You can easily find both of these integrals.