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DivGradCurl
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\int _{\sqrt{2}} ^2 \frac{1}{t^3 \sqrt{t^2 - 1}} \: dt = \frac{1}{24}\left( -6 + 3\sqrt{3} + \pi \right)
I've tried to use the trigonometric substitution t=\sec \theta, but without success thus far.
t=\sec \theta \Rightarrow \frac{dt}{d\theta}=\sec \theta \tan \theta \Rightarrow dt = \sec \theta \tan \theta \: d\theta
\int _{\sqrt{2}} ^{2} \frac{1}{t^3 \sqrt{t^2 - 1}} \: dt=\int _{\sec \sqrt{2}} ^{\sec 2} \frac{\sec \theta \tan \theta}{\sec ^3 \theta \sqrt{\sec ^2 \theta - 1}} \: d\theta
\int _{\sec \sqrt{2}} ^{\sec 2} \frac{\sec \theta \tan \theta}{\sec ^3 \theta \sqrt{\sec ^2 \theta - 1}} \: d\theta = \int _{\sec \sqrt{2}} ^{\sec 2} \frac{\tan \theta}{\sec ^2 \theta \sqrt{\tan ^2 \theta}} \: d\theta
\int _{\sec \sqrt{2}} ^{\sec 2} \frac{\tan \theta}{\sec ^2 \theta \sqrt{\tan ^2}} \: d\theta = \int _{\sec \sqrt{2}} ^{\sec 2} \frac{d\theta}{\sec ^2 \theta} = \int _{\sec \sqrt{2}} ^{\sec 2} \cos ^2 \theta \: d\theta
\int _{\sec \sqrt{2}} ^{\sec 2} \cos ^2 \theta \: d\theta = \frac{1}{2} \int _{\sec \sqrt{2}} ^{\sec 2} (1+\cos 2\theta) \: d\theta = \frac{1}{2} \int _{\sec \sqrt{2}} ^{\sec 2} \: d\theta + \frac{1}{2} \int _{\sec \sqrt{2}} ^{\sec 2} \cos 2\theta \: d\theta
Consider the following
\int \cos 2x \: dx
u=2x \Rightarrow \frac{du}{dx}= 2 \Rightarrow dx = \frac{du}{2}
\int \cos 2x \: dx = \frac{1}{2} \int \cos u \: du = \frac{1}{2} \sin u + \mathrm{C} = \frac{1}{2} \sin 2x + \mathrm{C} = \sin x \cos x + \mathrm{C}
Then, we get
\frac{1}{2} \int _{\sec \sqrt{2}} ^{\sec 2} \: d\theta + \frac{1}{2} \int _{\sec \sqrt{2}} ^{\sec 2} \cos 2\theta \: d\theta = \left. \frac{1}{2} \theta + \frac{1}{2} \sin \theta \cos \theta \right] _{\sec \sqrt{2}} ^{\sec 2}
which is wrong.
Any help is highly appreciated.
\int _{\sqrt{2}} ^2 \frac{1}{t^3 \sqrt{t^2 - 1}} \: dt = \frac{1}{24}\left( -6 + 3\sqrt{3} + \pi \right)
I've tried to use the trigonometric substitution t=\sec \theta, but without success thus far.
t=\sec \theta \Rightarrow \frac{dt}{d\theta}=\sec \theta \tan \theta \Rightarrow dt = \sec \theta \tan \theta \: d\theta
\int _{\sqrt{2}} ^{2} \frac{1}{t^3 \sqrt{t^2 - 1}} \: dt=\int _{\sec \sqrt{2}} ^{\sec 2} \frac{\sec \theta \tan \theta}{\sec ^3 \theta \sqrt{\sec ^2 \theta - 1}} \: d\theta
\int _{\sec \sqrt{2}} ^{\sec 2} \frac{\sec \theta \tan \theta}{\sec ^3 \theta \sqrt{\sec ^2 \theta - 1}} \: d\theta = \int _{\sec \sqrt{2}} ^{\sec 2} \frac{\tan \theta}{\sec ^2 \theta \sqrt{\tan ^2 \theta}} \: d\theta
\int _{\sec \sqrt{2}} ^{\sec 2} \frac{\tan \theta}{\sec ^2 \theta \sqrt{\tan ^2}} \: d\theta = \int _{\sec \sqrt{2}} ^{\sec 2} \frac{d\theta}{\sec ^2 \theta} = \int _{\sec \sqrt{2}} ^{\sec 2} \cos ^2 \theta \: d\theta
\int _{\sec \sqrt{2}} ^{\sec 2} \cos ^2 \theta \: d\theta = \frac{1}{2} \int _{\sec \sqrt{2}} ^{\sec 2} (1+\cos 2\theta) \: d\theta = \frac{1}{2} \int _{\sec \sqrt{2}} ^{\sec 2} \: d\theta + \frac{1}{2} \int _{\sec \sqrt{2}} ^{\sec 2} \cos 2\theta \: d\theta
Consider the following
\int \cos 2x \: dx
u=2x \Rightarrow \frac{du}{dx}= 2 \Rightarrow dx = \frac{du}{2}
\int \cos 2x \: dx = \frac{1}{2} \int \cos u \: du = \frac{1}{2} \sin u + \mathrm{C} = \frac{1}{2} \sin 2x + \mathrm{C} = \sin x \cos x + \mathrm{C}
Then, we get
\frac{1}{2} \int _{\sec \sqrt{2}} ^{\sec 2} \: d\theta + \frac{1}{2} \int _{\sec \sqrt{2}} ^{\sec 2} \cos 2\theta \: d\theta = \left. \frac{1}{2} \theta + \frac{1}{2} \sin \theta \cos \theta \right] _{\sec \sqrt{2}} ^{\sec 2}
which is wrong.
Any help is highly appreciated.
