Question

Solve this integration. I will mark u as BRAINLIEST...

F = $\int\limits^\alpha _\beta {sinx} \, dx$

Answer 1

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shreyvardhan08

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Answer:

we have to solve integral \int\limits^{\pi}_0{\frac{x}{1-cos\alpha sinx}}\,dx

first apply application : when \int\limits^a_b{f(x)}\,dx

then, f(x) = f(a + b - x)

so, I = \int\limits^{\pi}_0{\frac{x}{1-cos\alpha sinx}}\,dx=\int\limits^{\pi}_0{\frac{\pi-x}{1-cos\alpha sin(\pi-x)}}\,dx

so, 2I = \int\limits^{\pi}_0{\frac{x}{1-cos\alpha sinx}}\,dx+\int\limits^{\pi}_0{\frac{\pi-x}{1-cos\alpha sinx}}\,dx

⇒2I = \int\limits^{\pi}_0{\frac{\pi}{1-cos\alpha sinx}}\,dx

now putting, sinx = 2tan(x/2)/(1 + tan²(x/2))

so, 2I = \pi\int\limits^{\pi}_0{\frac{1}{1-cos\alpha\frac{2tan(x/2)}{1+tan^2(x/2)}}}\,dx

⇒2I/π = \int\limits^{\pi}_0{\frac{sec^2(x/2)}{1+tan^2(x/2)-cos\alpha 2tan(x/2)}}\,dx

[ because, 1 + tan²(x/2) = sec²(x/2)]

now putting, tan(x/2) = t

so, sec²(x/2)(1/2) dx = dt

⇒sec²(x/2) dx = 2dt

upper limit : tan(π) → ∞

lower limits : tan(0) → 0

⇒2I/π = \int\limits^{\infty}_0{\frac{2dt}{1+t^2-2cos\alpha t}}

1 + t² - 2cosα t = (t - cosα)² - (cos²α- 1)

= (t - cosα)² + sin²α

so, I/π = \int\limits^{\infty}_0{\frac{dt}{(t-cos\alpha)^2+sin^2\alpha}}

= \frac{1}{sin\alpha}\left[tan^{-1}\frac{1-cos\alpha}{sin\alpha}\right]^{\infty}_0

so, I = \frac{\pi}{sin\alpha}(\pi-\alpha)we have to solve integral \int\limits^{\pi}_0{\frac{x}{1-cos\alpha sinx}}\,dx

first apply application : when \int\limits^a_b{f(x)}\,dx

then, f(x) = f(a + b - x)

so, I = \int\limits^{\pi}_0{\frac{x}{1-cos\alpha sinx}}\,dx=\int\limits^{\pi}_0{\frac{\pi-x}{1-cos\alpha sin(\pi-x)}}\,dx

so, 2I = \int\limits^{\pi}_0{\frac{x}{1-cos\alpha sinx}}\,dx+\int\limits^{\pi}_0{\frac{\pi-x}{1-cos\alpha sinx}}\,dx

⇒2I = \int\limits^{\pi}_0{\frac{\pi}{1-cos\alpha sinx}}\,dx

now putting, sinx = 2tan(x/2)/(1 + tan²(x/2))

so, 2I = \pi\int\limits^{\pi}_0{\frac{1}{1-cos\alpha\frac{2tan(x/2)}{1+tan^2(x/2)}}}\,dx

⇒2I/π = \int\limits^{\pi}_0{\frac{sec^2(x/2)}{1+tan^2(x/2)-cos\alpha 2tan(x/2)}}\,dx

[ because, 1 + tan²(x/2) = sec²(x/2)]

now putting, tan(x/2) = t

so, sec²(x/2)(1/2) dx = dt

⇒sec²(x/2) dx = 2dt

upper limit : tan(π) → ∞

lower limits : tan(0) → 0

⇒2I/π = \int\limits^{\infty}_0{\frac{2dt}{1+t^2-2cos\alpha t}}

1 + t² - 2cosα t = (t - cosα)² - (cos²α- 1)

= (t - cosα)² + sin²α

so, I/π = \int\limits^{\infty}_0{\frac{dt}{(t-cos\alpha)^2+sin^2\alpha}}

= \frac{1}{sin\alpha}\left[tan^{-1}\frac{1-cos\alpha}{sin\alpha}\right]^{\infty}_0

so, I = \frac{\pi}{sin\alpha}(\pi-\alpha)we have to solve integral \int\limits^{\pi}_0{\frac{x}{1-cos\alpha sinx}}\,dx

first apply application : when \int\limits^a_b{f(x)}\,dx

then, f(x) = f(a + b - x)

so, I = \int\limits^{\pi}_0{\frac{x}{1-cos\alpha sinx}}\,dx=\int\limits^{\pi}_0{\frac{\pi-x}{1-cos\alpha sin(\pi-x)}}\,dx

so, 2I = \int\limits^{\pi}_0{\frac{x}{1-cos\alpha sinx}}\,dx+\int\limits^{\pi}_0{\frac{\pi-x}{1-cos\alpha sinx}}\,dx

⇒2I = \int\limits^{\pi}_0{\frac{\pi}{1-cos\alpha sinx}}\,dx

now putting, sinx = 2tan(x/2)/(1 + tan²(x/2))

so, 2I = \pi\int\limits^{\pi}_0{\frac{1}{1-cos\alpha\frac{2tan(x/2)}{1+tan^2(x/2)}}}\,dx

⇒2I/π = \int\limits^{\pi}_0{\frac{sec^2(x/2)}{1+tan^2(x/2)-cos\alpha 2tan(x/2)}}\,dx

[ because, 1 + tan²(x/2) = sec²(x/2)]

now putting, tan(x/2) = t

so, sec²(x/2)(1/2) dx = dt

⇒sec²(x/2) dx = 2dt

upper limit : tan(π) → ∞

lower limits : tan(0) → 0

⇒2I/π = \int\limits^{\infty}_0{\frac{2dt}{1+t^2-2cos\alpha t}}

1 + t² - 2cosα t = (t - cosα)² - (cos²α- 1)

= (t - cosα)² + sin²α

so, I/π = \int\limits^{\infty}_0{\frac{dt}{(t-cos\alpha)^2+sin^2\alpha}}

= \frac{1}{sin\alpha}\left[tan^{-1}\frac{1-cos\alpha}{sin\alpha}\right]^{\infty}_0

so, I = \frac{\pi}{sin\alpha}(\pi-\alpha)we have to solve integral \int\limits^{\pi}_0{\frac{x}{1-cos\alpha sinx}}\,dx

first apply application : when \int\limits^a_b{f(x)}\,dx

then, f(x) = f(a + b - x)

so, I = \int\limits^{\pi}_0{\frac{x}{1-cos\alpha sinx}}\,dx=\int\limits^{\pi}_0{\frac{\pi-x}{1-cos\alpha sin(\pi-x)}}\,dx

so, 2I = \int\limits^{\pi}_0{\frac{x}{1-cos\alpha sinx}}\,dx+\int\limits^{\pi}_0{\frac{\pi-x}{1-cos\alpha sinx}}\,dx

⇒2I = \int\limits^{\pi}_0{\frac{\pi}{1-cos\alpha sinx}}\,dx

now putting, sinx = 2tan(x/2)/(1 + tan²(x/2))

so, 2I = \pi\int\limits^{\pi}_0{\frac{1}{1-cos\alpha\frac{2tan(x/2)}{1+tan^2(x/2)}}}\,dx

⇒2I/π = \int\limits^{\pi}_0{\frac{sec^2(x/2)}{1+tan^2(x/2)-cos\alpha 2tan(x/2)}}\,dx

[ because, 1 + tan²(x/2) = sec²(x/2)]

now putting, tan(x/2) = t

so, sec²(x/2)(1/2) dx = dt

⇒sec²(x/2) dx = 2dt

upper limit : tan(π) → ∞

lower limits : tan(0) → 0

⇒2I/π = \int\limits^{\infty}_0{\frac{2dt}{1+t^2-2cos\alpha t}}

1 + t² - 2cosα t = (t - cosα)² - (cos²α- 1)

= (t - cosα)² + sin²α

so, I/π = \int\limits^{\infty}_0{\frac{dt}{(t-cos\alpha)^2+sin^2\alpha}}

= \frac{1}{sin\alpha}\left[tan^{-1}\frac{1-cos\alpha}{sin\alpha}\right]^{\infty}_0

so, I = \frac{\pi}{sin\alpha}(\pi-\alpha)

Step-by-step explanation:

Answer 2

Step-by-step explanation:

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