Question

Integrate $\int \frac{1}{1+\sin x}dx$ ?

Answer 1

Solve the given integral

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

Multiplying numerator and denominator by $\left(1-\sin x\right)$ we get ,

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

We know that,

$$\begin{gathered} {\sin }^{2}x+{\cos }^{2}x=1 \\ \Rightarrow {\cos }^{2}x=1-{\sin }^{2}x \end{gathered}$$

Now,

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

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

$$\begin{gathered} =\int \left(se{c}^{2}x-\tan xsecx\right)dx \\ =\tan x-secx+C \end{gathered}$$

Hence, Integral $\int \frac{1}{1+\sin x}dx$ is Integral $\tan x-secx+C$.