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Find the integral integral_sqrt(2)2 (1)/(t3 sqrt(t2-1)) dt using trigonometric substitution.

Mathematics

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Answer

π24+38−14\frac{\pi}{24} + \frac{\sqrt{3}}{8} - \frac{1}{4}

Step-by-step solution

5 steps

To find the integral ∫221t3t2−1dt\int_{\sqrt{2}}^{2} \frac{1}{t^3 \sqrt{t^2-1}} dt, we will use trigonometric substitution.

  1. Apply trigonometric substitution.

    Let t=sec⁡θt = \sec \theta.
    Then dt=sec⁡θtan⁡θdθdt = \sec \theta \tan \theta d\theta.
    And t2−1=sec⁡2θ−1=tan⁡2θ=tan⁡θ\sqrt{t^2-1} = \sqrt{\sec^2 \theta - 1} = \sqrt{\tan^2 \theta} = \tan \theta (assuming θ∈(0,π/2)\theta \in (0, \pi/2)).

  2. Change the limits of integration.

    When t=2t = \sqrt{2}:
    sec⁡θ=2  ⟹  cos⁡θ=12  ⟹  θ=π4\sec \theta = \sqrt{2} \implies \cos \theta = \frac{1}{\sqrt{2}} \implies \theta = \frac{\pi}{4}.
    When t=2t = 2:
    sec⁡θ=2  ⟹  cos⁡θ=12  ⟹  θ=π3\sec \theta = 2 \implies \cos \theta = \frac{1}{2} \implies \theta = \frac{\pi}{3}.

  3. Substitute into the integral and simplify.

    ∫221t3t2−1dt=∫π/4π/31(sec⁡θ)3(tan⁡θ)(sec⁡θtan⁡θ)dθ\int_{\sqrt{2}}^{2} \frac{1}{t^3 \sqrt{t^2-1}} dt = \int_{\pi/4}^{\pi/3} \frac{1}{(\sec \theta)^3 (\tan \theta)} (\sec \theta \tan \theta) d\theta =∫π/4π/3sec⁡θtan⁡θsec⁡3θtan⁡θdθ= \int_{\pi/4}^{\pi/3} \frac{\sec \theta \tan \theta}{\sec^3 \theta \tan \theta} d\theta =∫π/4π/31sec⁡2θdθ= \int_{\pi/4}^{\pi/3} \frac{1}{\sec^2 \theta} d\theta =∫π/4π/3cos⁡2θdθ= \int_{\pi/4}^{\pi/3} \cos^2 \theta d\theta

  4. Use the power-reducing identity for cos⁡2θ\cos^2 \theta.

    Recall that cos⁡2θ=1+cos⁡(2θ)2\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}.

    =∫π/4π/31+cos⁡(2θ)2dθ= \int_{\pi/4}^{\pi/3} \frac{1 + \cos(2\theta)}{2} d\theta =12∫π/4π/3(1+cos⁡(2θ))dθ= \frac{1}{2} \int_{\pi/4}^{\pi/3} (1 + \cos(2\theta)) d\theta

  5. Integrate and evaluate the definite integral.

    =12[θ+12sin⁡(2θ)]π/4π/3= \frac{1}{2} \left[ \theta + \frac{1}{2}\sin(2\theta) \right]_{\pi/4}^{\pi/3} =12[(π3+12sin⁡(2⋅π3))−(π4+12sin⁡(2⋅π4))]= \frac{1}{2} \left[ \left( \frac{\pi}{3} + \frac{1}{2}\sin\left(2 \cdot \frac{\pi}{3}\right) \right) - \left( \frac{\pi}{4} + \frac{1}{2}\sin\left(2 \cdot \frac{\pi}{4}\right) \right) \right] =12[(π3+12sin⁡(2π3))−(π4+12sin⁡(π2))]= \frac{1}{2} \left[ \left( \frac{\pi}{3} + \frac{1}{2}\sin\left(\frac{2\pi}{3}\right) \right) - \left( \frac{\pi}{4} + \frac{1}{2}\sin\left(\frac{\pi}{2}\right) \right) \right] =12[(π3+12⋅32)−(π4+12⋅1)]= \frac{1}{2} \left[ \left( \frac{\pi}{3} + \frac{1}{2} \cdot \frac{\sqrt{3}}{2} \right) - \left( \frac{\pi}{4} + \frac{1}{2} \cdot 1 \right) \right] =12[π3+34−π4−12]= \frac{1}{2} \left[ \frac{\pi}{3} + \frac{\sqrt{3}}{4} - \frac{\pi}{4} - \frac{1}{2} \right] =12[(4π−3π12)+34−12]= \frac{1}{2} \left[ \left(\frac{4\pi - 3\pi}{12}\right) + \frac{\sqrt{3}}{4} - \frac{1}{2} \right] =12[π12+34−12]= \frac{1}{2} \left[ \frac{\pi}{12} + \frac{\sqrt{3}}{4} - \frac{1}{2} \right] =π24+38−14= \frac{\pi}{24} + \frac{\sqrt{3}}{8} - \frac{1}{4}

    The final answer is π24+38−14\frac{\pi}{24} + \frac{\sqrt{3}}{8} - \frac{1}{4}.

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