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Exercise 7.5 · Q17

Q.Integrate the following function: cos⁡x(1−sin⁡x)(2−sin⁡x)\frac{\cos x}{(1 - \sin x)(2 - \sin x)} [Hint : Put sin⁡x=t\sin x = t]

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Substituting sin⁡x=t\sin x = t turns the integral into ∫dt(1−t)(2−t)\int \dfrac{dt}{(1-t)(2-t)}; partial fractions give log⁡∣2−sin⁡x1−sin⁡x∣+C\log\left|\dfrac{2-\sin x}{1-\sin x}\right| + C.

Substitute. Let t=sin⁡xt = \sin x, so dt=cos⁡x dxdt = \cos x\,dx:

∫cos⁡x(1−sin⁡x)(2−sin⁡x) dx=∫dt(1−t)(2−t).\int \frac{\cos x}{(1-\sin x)(2-\sin x)}\,dx = \int \frac{dt}{(1-t)(2-t)}.

Partial fractions. 1(1−t)(2−t)=A1−t+B2−t\dfrac{1}{(1-t)(2-t)} = \dfrac{A}{1-t} + \dfrac{B}{2-t} with 1=A(2−t)+B(1−t)1 = A(2-t) + B(1-t). Setting t=1t=1 gives A=1A=1; setting t=2t=2 gives B=−1B=-1. So

1(1−t)(2−t)=11−t−12−t.\frac{1}{(1-t)(2-t)} = \frac{1}{1-t} - \frac{1}{2-t}.

Integrate. Since ∫dta−t=−log⁡∣a−t∣\int \dfrac{dt}{a-t} = -\log|a-t|, …

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