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Integrals

Short Answer Type

251.

Find space integral fraction numerator dx over denominator straight x squared space plus 4 straight x space plus 8 end fraction

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252.

Find $integral fraction numerator 2 straight x over denominator left parenthesis straight x squared plus 1 right parenthesis left parenthesis straight x squared plus 2 right parenthesis squared end fraction dx$

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253.

Evaluate: $integral subscript 0 superscript straight pi fraction numerator straight x space sin space straight x over denominator 1 plus cos squared x end fraction dx$

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254.

Evaluate: $integral subscript 0 superscript 3 divided by 2 end superscript vertical line space straight x space sin space πx vertical line space dx$

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255.

Evaluate: $\int \frac{\cos 2 x + 2 \sin^{2} x}{\cos^{2} x} dx$

256.
Find: $\int \frac{2 \cos x}{(1 - \sin x) (1 + \sin^{2} x)} dx$

$Let I = \int \frac{2 \cos x}{(1 - \sin x) (1 + \sin^{2} x)} dx Let \sin x = t \cos dx = dt ∴ I = \int \frac{2}{(1 - t) (1 + t^{2})} dt consider \frac{2}{(1 - t) (1 + t^{2})} = \frac{A}{1 - t} + \frac{Bt + C}{t^{2} + 1} = \frac{A (t^{2} + 1) + (Bt + C) (1 - t)}{(1 - t) (t^{2} + 1)} ∴ 2 {At}^{2} + A + Bt + C - {Bt}^{2} - Ct = (A - B) t^{2} + (B - C) t + (A + C) ∴ A - B = 0, B - C = 0 A + C = 2 A = 1, B = 1, C = 1 ∴ I = \int (\frac{1}{1 - t} + \frac{2 t}{2 (t^{2} + 1)} + \frac{1}{t^{2} + 1}) dt = - \log | 1 - t | + \frac{1}{2} \log | t^{2} + 1 | + \tan^{- 1} (t) + C = \frac{1}{2} \log |\frac{t^{2} + 1}{(1 - t)^{2}}| + \tan^{- 1} (t) + C = \frac{1}{2} \log |\frac{\sin^{2} x + 1}{(1 - \sin x)^{2}}| + \tan^{- 1} (\sin x) + C$