How to integrate 1/(1+sinx)?

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How to integrate 1/(1+sinx)?

I'm currently learning Trigonometric Functions and need help with solving two questions:

$\int$ $int$ $\frac{1}{1 + sin x}$ $1/(1+sinx)$ $d x$ $dx$ = ?

$\int$ $int$ $\frac{1}{1 + cos 2 x}$ $1/(1+cos2x)$ $d x$ $dx$ = ?

Thanks!

1 Answer

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Answer 1 · 2017-10-02T07:40:00

$\int \frac{1}{1 + sin x} d x = tan x - sec x$ $int1/(1+sinx)dx=tanx-secx$ and $\int \frac{1}{1 + cos 2 x} d x = \frac{1}{2} tan x$ $int1/(1+cos2x)dx=1/2tanx$

Explanation:

$\int \frac{1}{1 + sin x} d x$ $int1/(1+sinx)dx$

= $\int \frac{1 - sin x}{1 - {sin}^{2} x} d x$ $int(1-sinx)/(1-sin^2x)dx$

= $\int \frac{1 - sin x}{{cos}^{2} x} d x$ $int(1-sinx)/cos^2xdx$

= $\int ({sec}^{2} x - tan x sec x) d x$ $int(sec^2x-tanxsecx)dx$

= $tan x - sec x$ $tanx-secx$

$\int \frac{1}{1 + cos 2 x} d x$ $int1/(1+cos2x)dx$

= $\int \frac{1}{1 + 2 {cos}^{2} x - 1} d x$ $int1/(1+2cos^2x-1)dx$

= $\int \frac{1}{2 {cos}^{2} x} d x$ $int1/(2cos^2x)dx$

= $\frac{1}{2} \int {sec}^{2} x d x$ $1/2intsec^2xdx$

= $\frac{1}{2} tan x$ $1/2tanx$

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How to integrate 1/(1+sinx)?

I'm currently learning Trigonometric Functions and need help with solving two questions:

$\int$ $int$ $\frac{1}{1 + sin x}$ $1/(1+sinx)$ $d x$ $dx$ = ?

$\int$ $int$ $\frac{1}{1 + cos 2 x}$ $1/(1+cos2x)$ $d x$ $dx$ = ?

Thanks!

1 Answer

Explanation:

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How to integrate 1/(1+sinx)?

I'm currently learning Trigonometric Functions and need help with solving two questions: ∫int 11+sinx1/(1+sinx) dxdx = ? ∫int 11+cos2x1/(1+cos2x) dxdx = ? Thanks!

1 Answer

Explanation:

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Impact of this question

I'm currently learning Trigonometric Functions and need help with solving two questions:

$\int$ $int$ $\frac{1}{1 + sin x}$ $1/(1+sinx)$ $d x$ $dx$ = ?

$\int$ $int$ $\frac{1}{1 + cos 2 x}$ $1/(1+cos2x)$ $d x$ $dx$ = ?

Thanks!