Question

Using $t = \sec(\theta)$, we get $\int_{\sqrt{2}}^{2} \frac{1}{t^3 \sqrt{t^2 - 1}} dt = \int_{\pi/4}^{\pi/3} \frac{\text{____}}{\sec^3(\theta)\tan(\theta)} \sec(\theta)\tan(\theta) d\theta$. This can be further simplified to $\int_{\pi/4}^{\pi/3} \frac{\text{____}}{\sec^2(\theta)} d\theta$, or simply $\int_{\pi/4}^{\pi/3} \cos^2(\theta) d\theta$.

          Using $t = \sec(\theta)$, we get
$\int_{\sqrt{2}}^{2} \frac{1}{t^3 \sqrt{t^2 - 1}} dt = \int_{\pi/4}^{\pi/3} \frac{\text{____}}{\sec^3(\theta)\tan(\theta)} \sec(\theta)\tan(\theta) d\theta$.
This can be further simplified to
$\int_{\pi/4}^{\pi/3} \frac{\text{____}}{\sec^2(\theta)} d\theta$,
or simply
$\int_{\pi/4}^{\pi/3} \cos^2(\theta) d\theta$.

Using t = sec(θ), we get
∫√(2)^2(1)/(t^3 √(t^2 - 1)) dt = ∫π/4^π/3()/(sec^3(θ)tan(θ))sec(θ)tan(θ) dθ.
This can be further simplified to
∫π/4^π/3()/(sec^2(θ)) dθ,
or simply
∫π/4^π/3cos^2(θ) dθ.

Added by Patricia G.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

06/19/2023

Step 1

Step 1: Substitute $t = sec(\theta)$ and $dt = sec(\theta)tan(\theta) d\theta$ in the integral. Show more…

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Using $t = sec(\theta)$, we get $$\int_{\sqrt{2}}^{2} \frac{1}{t^3 \sqrt{t^2 - 1}} dt = \int_{\pi/4}^{\pi/3} \boxed{ } \sec(\theta) \tan(\theta) d\theta.$$ This can be further simplified to $$\int_{\pi/4}^{\pi/3} \boxed{ } \frac{d\theta}{\sec^2(\theta)},$$ or simply $$\int_{\pi/4}^{\pi/3} \cos^2(\theta) d\theta.$$