Apply Weierstrass substitution: Use the Weierstrass substitution to simplify the integral. Let t = tan(x/2), then sin(x) = 2t/(1 + t^2) and dx = 2/(1 + t^2) dt.Substitute sin(x) and dx: Substitute sin(x) and dx in the integral. int(1)/(1 - sin x)dx = int(1)/(1 - 2t/(1 + t^2)) * (2/(1 + t^2))dtSimplify the integrand: Simplify the integrand. int(1)/(1 - 2t/(1 + t^2)) * (2/(1 + t^2))dt = int((1 + t^2)/(1 + t^2 - 2t)) * (2/(1 + t^2))dt = int(2/(1 - t^2))dtDecompose into partial fractions: Decompose the integrand into partial fractions. 2/(1 - t^2) can be written as A/(1 - t) + B/(1 + t). Solving for A and B, we get A = 1 and B = 1.Write in terms of partial fractions: Write the integral in terms of the partial fractions. int(2/(1 - t^2))dt = int(1/(1 - t))dt + int(1/(1 + t))dtIntegrate each term separately: Integrate each term separately. int(1/(1 - t))dt = -ln|1 - t| + C1 int(1/(1 + t))dt = ln|1 + t| + C2Combine the two integrals: Combine the two integrals. int(2/(1 - t^2))dt = -ln|1 - t| + ln|1 + t| + C = ln|1 + t| - ln|1 - t| + CSubstitute back for t: Substitute back for t = tan(x/2). ln|1 + tan(x/2)| - ln|1 - tan(x/2)| + C = ln|(1 + tan(x/2))/(1 - tan(x/2))| + CSimplify using trigonometric identities: Simplify the expression using trigonometric identities. ln|(1 + tan(x/2))/(1 - tan(x/2))| + C = ln|((1 + tan(x/2))^2)/((1 - tan(x/2))(1 + tan(x/2)))| + C = ln|((1 + tan(x/2))^2)/(1 - tan^2(x/2))| + C = ln|(1 + sin(x))/(1 - sin(x))| + CCheck final expression: Check if the final expression is correct. The final expression ln|(1 + sin(x))/(1 - sin(x))| + C is the antiderivative of 1/(1 - sin x), which answers the question prompt.

Resources

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

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

View step-by-step help

Home

Math Problems

Calculus

Evaluate definite integrals using the chain rule

Full solution

\int \frac{1}{1-\sin x} \, dx

Apply Weierstrass substitution: Use the Weierstrass substitution to simplify the integral. Let $t = \tan(\frac{x}{2})$ , then $\sin(x) = \frac{2t}{1 + t^2}$ and $dx = \frac{2}{1 + t^2} dt$ .
Substitute $\sin(x)$ and $dx$ : Substitute $\sin(x)$ and $dx$ in the integral. $\int\frac{1}{1 - \sin x}\,dx = \int\frac{1}{1 - \frac{2t}{1 + t^2}} \cdot \frac{2}{1 + t^2}\,dt$
Simplify the integrand: Simplify the integrand. $\int\frac{1}{1 - \frac{2t}{1 + t^2}} \cdot \frac{2}{1 + t^2}\,dt = \int\frac{1 + t^2}{1 + t^2 - 2t} \cdot \frac{2}{1 + t^2}\,dt = \int\frac{2}{1 - t^2}\,dt$
Decompose into partial fractions: Decompose the integrand into partial fractions. $\newline$ $\frac{2}{1 - t^2}$ can be written as $\frac{A}{1 - t} + \frac{B}{1 + t}$ . $\newline$ Solving for $A$ and $B$ , we get $A = 1$ and $B = 1$ .
Write in terms of partial fractions: Write the integral in terms of the partial fractions. \int\frac{$2$}{$1$ - t^$2$}\,dt = \int\frac{$1$}{$1$ - t}\,dt + \int\frac{$1$}{$1$ + t}\,dt
Integrate each term separately: Integrate each term separately.\(\newline $\int\frac{1}{1 - t}\,dt = -\ln|1 - t| + C_1$ $\newline$ $\int\frac{1}{1 + t}\,dt = \ln|1 + t| + C_2$
Combine the two integrals: Combine the two integrals. $\int \frac{2}{1 - t^2}\,dt = -\ln|1 - t| + \ln|1 + t| + C = \ln|1 + t| - \ln|1 - t| + C$
Substitute back for t: Substitute back for t = $\tan(\frac{x}{2})$ . $\newline$ $\ln|1 + \tan(\frac{x}{2})| - \ln|1 - \tan(\frac{x}{2})| + C = \ln|\frac{(1 + \tan(\frac{x}{2}))}{(1 - \tan(\frac{x}{2}))}| + C$
Simplify using trigonometric identities: Simplify the expression using trigonometric identities. $\newline$ \ln\left|\frac{$1$ + \tan(\frac{x}{$2$})}{$1$ - \tan(\frac{x}{$2$})}\right| + C = \ln\left|\frac{($1$ + \tan(\frac{x}{$2$}))^$2$}{($1$ - \tan(\frac{x}{$2$}))($1$ + \tan(\frac{x}{$2$}))}\right| + C\(\newline= \ln\left|\frac{( $1$ + \tan(\frac{x}{ $2$ }))^ $2$ }{ $1$ - \tan^ $2$ (\frac{x}{ $2$ })}\right| + C $\newline$ = \ln\left|\frac{ $1$ + \sin(x)}{ $1$ - \sin(x)}\right| + C
Check final expression: Check if the final expression is correct. $\newline$ The final expression $\ln\left|\frac{(1 + \sin(x))}{(1 - \sin(x))}\right| + C$ is the antiderivative of $\frac{1}{(1 - \sin x)}$ , which answers the question prompt.