Question

Question :

$\displaystyle \int \sf\dfrac{cos2x - cos2\alpha}{cosx - cos\alpha}dx$

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Answer 1

$\;\qquad\quad\;\;\large{\odot\;\;\underline{\underline{\bf{\purple{Given\; Question}}}}}$

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$\displaystyle \int \sf\dfrac{cos2x - cos2\alpha}{cosx - cos\alpha}dx$

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$\;\qquad\quad\;\;\large{\odot\;\;\underline{\underline{\bf{\blue{Required\; Solution}}}}}$

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$\displaystyle \sf \bold{2(sin \: x \: + x \: cos \alpha ) + c}$

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$\;\qquad\quad\;\;\large{\odot\;\;\underline{\underline{\bf{\green{Step \; By \; Step\; Solution}}}}}$

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$\displaystyle \begin{gathered}\\\;\;\sf{:\rightarrow\;\int \sf\dfrac{cos2x - cos2\alpha}{cosx - cos\alpha}dx}\end{gathered}$

$\displaystyle \begin{gathered}\\\;\;\sf{:\longrightarrow\;\int \sf\dfrac{(2cos^{2} x -1) - (2 cos ^{2} \alpha - 1)}{cosx - cos\alpha}dx}\end{gathered}$

$\displaystyle \begin{gathered}\\\;\;\sf{:\longrightarrow\;\int \sf\dfrac{2cos^{2} x - 1 - 2 cos ^{2} \alpha + 1}{cosx - cos\alpha}dx}\end{gathered}$

$\displaystyle \begin{gathered}\\\;\;\sf{:\longrightarrow\;\int \sf\dfrac{2cos^{2} x - 2 cos ^{2} \alpha + 1 - 1}{cosx - cos\alpha}dx}\end{gathered}$

$\displaystyle \begin{gathered}\\\;\;\sf{:\longrightarrow\;\int \sf\dfrac{2(cos^{2} x - 2 cos ^{2} \alpha ) }{cosx - cos\alpha}dx}\end{gathered}$

$\displaystyle \begin{gathered}\\\;\;\sf{:\longrightarrow\;2\int \sf\dfrac{cos^{2} x - 2 cos ^{2} \alpha }{cosx - cos\alpha}dx}\end{gathered}$

$\displaystyle \begin{gathered}\\\;\;\sf{:\longrightarrow\;2\int \sf\dfrac{(cos x - cos \alpha) (cos x + cos \alpha) }{(cosx - cos\alpha)}dx}\end{gathered}$

$\displaystyle \begin{gathered}\\\;\;\sf{:\longrightarrow\;2\int \sf{(cos x + cos \alpha) }dx}\end{gathered}$

$\displaystyle \begin{gathered}\\\;\;\sf{:\longrightarrow\;2(\int \sf{cos x \: dx + \int cos \alpha }dx)}\end{gathered}$

$\displaystyle \begin{gathered}\\\;\;\sf{:\longrightarrow\;2(\int \sf{cos x \: dx + cos \alpha \int1. \: }dx)}\end{gathered}$

$\displaystyle \Longrightarrow\sf \bold{2(sin \: x \: + x \: cos \alpha ) + c}$

Answer 2

EXPLANATION.

∫[cos2x - cos2α/cos x - cos α]dx.

As we know that,

⇒ cos2∅ = 2cos²∅ - 1.

put cos2x = cos2α = 2cos²∅ - 1.

⇒ ∫[(2cos²x - 1) - (2cos²α - 1)/cos x - cosα]dx.

⇒ ∫[2cos²x - 2cos²α - 1 + 1/cos x - cosα]dx.

⇒ ∫[2cos²x - 2cos²α/cos x - cosα]dx.

⇒ ∫[2 ( cos²x - cos²α)/cos x - cosα]dx.

⇒ ∫[2 ( cos x - cosα)(cos x + cosα)/(cos x - cosα)]dx.

⇒ ∫[2(cos x + cosα)dx.

⇒ 2[(∫cos x)dx + ∫(cosα)dx].

⇒ 2sinx + 2xcosα + c.

                                                                   

MORE INFORMATION.

(1) = Integration of a function.

⇒ ∫f(x)dx = Ф(x) + c ⇔ d/dx [Ф(x)] = f(x).

(2) = Basic Theorem on integration.

(1) = ∫Kf(x)dx = K∫f(x)dx + c.

(2) = ∫[f(x) ± g(x)]dx = ∫f(x)dx ± ∫g(x)dx + c.

(3) = d/dx [∫f(x)dx] = f(x).

(4) = ∫[d/dx f(x)]dx = f(x).