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511sinxdx5\int \frac{1}{1-\sin x}\,dx\Rightarrow

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Q. 511sinxdx5\int \frac{1}{1-\sin x}\,dx\Rightarrow
  1. Trig Identity Simplification: We are given the integral to solve: \newline51sin(x)dx\int \frac{5}{1 - \sin(x)} \, dx\newlineTo solve this integral, we can use a trigonometric identity to simplify the integrand. The identity we will use is:\newlinesin2(x)+cos2(x)=1\sin^2(x) + \cos^2(x) = 1\newlineThis identity allows us to express sin(x)\sin(x) in terms of cos(x)\cos(x), which can be useful for simplification.
  2. Pythagorean Identity: We can rewrite the denominator 1sin(x)1 - \sin(x) using the Pythagorean identity by adding and subtracting cos2(x)\cos^2(x) to the denominator:\newline1sin(x)=1sin2(x)+cos2(x)cos2(x)1 - \sin(x) = 1 - \sin^2(x) + \cos^2(x) - \cos^2(x)\newline=(1sin2(x))+cos2(x)cos2(x)= (1 - \sin^2(x)) + \cos^2(x) - \cos^2(x)\newline=cos2(x)+(1sin2(x))cos2(x)= \cos^2(x) + (1 - \sin^2(x)) - \cos^2(x)\newline=cos2(x)+cos2(x)cos2(x)= \cos^2(x) + \cos^2(x) - \cos^2(x)\newline=cos2(x)= \cos^2(x)\newlineNow we have:\newline51sin(x)dx=5cos2(x)dx\int \frac{5}{1 - \sin(x)} dx = \int \frac{5}{\cos^2(x)} dx
  3. Weierstrass Substitution: To simplify the integral, we can use the Weierstrass substitution, which involves the substitution:\newlinet=tan(x2)t = \tan(\frac{x}{2})\newlineThen we have:\newlinesin(x)=2t1+t2\sin(x) = \frac{2t}{1 + t^2}\newlinecos(x)=1t21+t2\cos(x) = \frac{1 - t^2}{1 + t^2}\newlinedx=21+t2dtdx = \frac{2}{1 + t^2} dt\newlineNow we can rewrite the integral in terms of tt.
  4. Integrand Simplification: Substituting the expressions for sin(x)\sin(x), cos(x)\cos(x), and dxdx into the integral, we get:\newline51sin(x)dx=512t1+t221+t2dt\int \frac{5}{1 - \sin(x)} dx = \int \frac{5}{1 - \frac{2t}{1 + t^2}} \cdot \frac{2}{1 + t^2} dt\newlineNow we need to simplify the integrand.
  5. Standard Integral Formula: Simplify the integrand:\newline= 51+t21+t22t21+t2dt\int 5 \cdot \frac{1 + t^2}{1 + t^2 - 2t} \cdot \frac{2}{1 + t^2} \, dt\newline= 521+t22tdt\int 5 \cdot \frac{2}{1 + t^2 - 2t} \, dt\newline= 1012t+t2dt\int \frac{10}{1 - 2t + t^2} \, dt\newline= 10(t1)2dt\int \frac{10}{(t - 1)^2} \, dt\newlineNow we have a simpler integral to solve.
  6. Substitution for t: The integral 10(t1)2dt\int \frac{10}{(t - 1)^2} \, dt is a standard integral that can be solved using the formula:\newlinea(xb)2dx=a(xb)+C\int \frac{a}{(x - b)^2} \, dx = -\frac{a}{(x - b)} + C\newlinewhere aa is a constant, bb is the constant term in the denominator, and CC is the constant of integration.\newlineApplying this formula, we get:\newline10(t1)2dt=10(t1)+C\int \frac{10}{(t - 1)^2} \, dt = -\frac{10}{(t - 1)} + C
  7. Final Answer: Now we need to substitute back for tt to express the answer in terms of xx. Recall that t=tan(x2)t = \tan(\frac{x}{2}), so we have:\newline-\frac{\(10\)}{t - \(1\)} + C = -\frac{\(10\)}{\tan(\frac{x}{\(2\)}) - \(1\)} + C\(\newlineThis is the antiderivative in terms of xx.

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