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Integrate the following w.r.t. x : 1sinx+sin2x - Mathematics and Statistics

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Question

Integrate the following w.r.t. x : $\frac{1}{\sin x + \sin 2 x}$

Sum

Solution

Let I = $\int \frac{1}{\sin x + \sin 2 x} \cdot d x$

= $\int \frac{1}{\sin x + 2 \sin x \cos x} \cdot d x$

= $\int \frac{d x}{\sin x (1 + 2 \cos x)}$

= $\int \frac{\sin x \cdot d x}{\sin^{2} x (1 + 2 \cos x)}$

= $\int \frac{\sin \cdot d x}{(1 - \cos^{2} x) (1 + 2 \cos x)}$

= $\int \frac{\sin \cdot d x}{(1 - \cos x) (1 + \cos x) (1 + 2 \cos x)}$
Put cos x = t
∴ – sinx . dx = dt
∴ sinx .dx = – dt

∴ I = $\int \frac{- d t}{(1 - t) (1 + t) (1 + 2 t)}$

= $- \int \frac{d t}{(1 - t) (1 + t) (1 + 2 t)}$

Let $\frac{1}{(1 - t) (1 + t) (1 + 2 t)} = \frac{A}{1 - t} + \frac{B}{1 + t} + \frac{C}{1 + 2 t}$

∴ 1 = A(1 + t)(1 + 2t) + B(1 – t)(1 + 2t) + C(1 – t)(1 + t)
Putting 1 – t = 0, i.e. t = 1, we get
1 = A(2)(3) + B(0)(3) + C(0)(2)
∴ A = $\frac{1}{6}$
Putting 1 – t = 0, i.e. t = – 1, we get
1 = A(0)(– 1) + B(2)(– 1) + C(2)(0)
∴ B = $- \frac{1}{2}$
Putting 1 + 2t = 0, i.e. t = $- \frac{1}{2}$ , we get

1 = $A (0) + B (0) + C (\frac{3}{2}) (\frac{1}{2})$

∴ C = $\frac{4}{3}$

∴ $\frac{1}{(1 - t) (1 + t) (1 + 2 t)} = \frac{(\frac{1}{6})}{1 - t} + \frac{(\frac{- 1}{2})}{1 + t} + \frac{(\frac{4}{3})}{1 + 2 t}$

∴ I = $\int [\frac{(\frac{1}{6})}{1 - t} + \frac{(\frac{- 1}{2})}{1 + t} + \frac{(\frac{4}{3})}{1 + 2 t}] \cdot d t$

= $- \frac{1}{6} \int \frac{1}{1 - t} \cdot d t + \frac{1}{2} \int \frac{1}{1 + t} \cdot d t - \frac{4}{3} \int \frac{1}{1 + 2 t} \cdot d t$

= $- \frac{1}{6} \cdot \frac{\log | 1 - t |}{- 1} + \frac{1}{2} \log | 1 + t | - \frac{4}{3} \cdot \frac{\log | 1 + 2 t |}{2} + c$

= $- \frac{1}{6} \log | 1 - \cos x | + \frac{1}{2} \log | 1 + \cos x | - \frac{2}{3} \log | 1 + 2 \cos x | + c$

= $\frac{1}{2} \log | \cos x + 1 | + \frac{1}{6} \log | \cos x - 1 | - \frac{2}{3} \log | 2 \cos x + 1 | + c$ .

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Methods of Integration: Integration Using Partial Fractions

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Chapter 3: Indefinite Integration - Exercise 3.4 [Page 145]

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Balbharati Mathematics and Statistics 2 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board

Chapter 3 Indefinite Integration
Exercise 3.4 | Q 1.18 | Page 145

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