SATHEE: Indefinite Integration 1 Question 67

Indefinite Integration 1 Question 67

68. Evaluate $\int_{- π / 3}^{π / 3} \frac{π + 4 x^{3}}{2 - \cos | x | + \frac{π}{3}} d x$ .

$(2004, 4$ M)

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Solution:

Let $I = \int_{- π / 3}^{π / 3} \frac{π d x}{2 - \cos | x | + \frac{π}{3}} + 4 \int_{- π / 3}^{π / 3} \frac{x^{3} d x}{2 - \cos | x | + \frac{π}{3}}$

$\begin{matrix} Using \int_{- a}^{a} f (x) d x = \int_{0}^{a} f (x) d x, f (- x) = f (x) \\ ∴ I = 2 \int_{0}^{π / 3} \frac{π d x}{2 - \cos | x | + \frac{π}{3}} + 0 \\ ∵ \frac{x^{3} d x}{2 - \cos | x | + \frac{π}{3}} is odd \\ I = 2 π \int_{0}^{π / 3} \frac{d x}{2 - \cos (x + π / 3)} \end{matrix}$

Put $x + \frac{π}{3} = t \Rightarrow d x = d t$

$∴ I = 2 π \int_{π / 3}^{2 π / 3} \frac{d t}{2 - \cos t} = 2 π \int_{π / 3}^{2 π / 3} \frac{\sec^{2} \frac{t}{2} d t}{1 + 3 \tan^{2} \frac{t}{2}}$

Put $\tan \frac{t}{2} = u \Rightarrow \sec^{2} \frac{t}{2} d t = 2 d u$

$\begin{aligned} \Rightarrow & I = 2 π \int_{1 / \sqrt{3}}^{\sqrt{3}} \frac{2 d u}{1 + 3 u^{2}} = \frac{4 π}{3} {[\sqrt{3} \tan^{- 1} \sqrt{3} u]}_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \\ = \frac{4 π}{\sqrt{3}} (\tan^{- 1} 3 - \tan^{- 1} 1) = \frac{4 π}{\sqrt{3}} \tan^{- 1} \frac{1}{2} \\ ∴ & \int_{- π / 3}^{π / 3} \frac{π + 4 x^{3}}{2 - \cos | x | + \frac{π}{3}} d x = \frac{4 π}{\sqrt{3}} \tan^{- 1} \frac{1}{2} \end{aligned}$

Indefinite Integration 1 Question 67

68. Evaluate $\int_{- π / 3}^{π / 3} \frac{π + 4 x^{3}}{2 - \cos | x | + \frac{π}{3}} d x$ .

Solution:

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Indefinite Integration 1 Question 67

68. Evaluate ∫−π/3π/3π+4x32−cos⁡|x|+π3dx.

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68. Evaluate $\int_{- π / 3}^{π / 3} \frac{π + 4 x^{3}}{2 - \cos | x | + \frac{π}{3}} d x$ .