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

Long Answer Type

257.

$Evaluate : \int \frac{\sin x + \cos x}{16 + 9 \sin 2 x} dx$

258.

Evaluate: $\int_{1}^{3} (x^{2} + 3 x + e^{x}) dx$

Short Answer Type

259.

Evaluate: $\int \frac{x^{2}}{1 + x^{3}} dx$

260.
Evaluate: $\int_{0}^{4} \frac{dx}{1 + x^{2}} dx$

$\int_{0}^{4} \frac{dx}{1 + x^{2}} dx Let X = tanθ \Rightarrow θ = \tan^{- 1} X dx = \sec^{2} θ dθ When X = 0, θ = \tan^{- 1} (0) = 0 When X = 1, θ = \tan^{- 1} 1 = \frac{π}{4} ∴ \int_{0}^{4} \frac{dx}{1 + x^{2}} = \int_{0}^{\frac{π}{4}} \frac{\sec^{2} θ}{1 + \tan^{2} θ} dθ = \int_{0}^{\frac{π}{4}} \frac{\sec^{2} θ}{\sec^{2} θ} dθ = \int_{0}^{\frac{π}{4}} dθ = {[θ]}_{0}^{\frac{π}{4}} = [\frac{π}{4} - 0] = \frac{π}{4}$

Previous Year Papers

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Test Yourself

Short Answer Type

Long Answer Type

Short Answer Type

260. Evaluate: ∫04dx1 + x2 dx

260.
Evaluate: $\int_{0}^{4} \frac{dx}{1 + x^{2}} dx$