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$$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx$$

$$f(x) = (\tan \ x)^{\frac{2}{3}}, \ f'(x) = \frac{2}{3} \cdot (\tan \ x)^{-\frac{1}{3}} \cdot \sec^2x$$

$$\therefore \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx = \frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{2}{3}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx = \frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{f'(x)}{f(x)}dx$$

$$\therefore \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx = \frac{3}{2}[\ln(\tan \ x)^{\frac{2}{3}}]_{\frac{\pi}{4}}^{\frac{\pi}{3}} = \frac{3}{2}[\ln(\sqrt{3})^\frac{2}{3}-\ln(1)^\frac{2}{3}]=\frac{3}{2}[\ln(\sqrt{3})^\frac{2}{3}] = \bf \color{red}{\ln(\sqrt{3})}$$

However, the answer given is not this value, but instead $= \frac{3\sqrt[3]{3}-3}{2} \approx 0.663... $ while $\ln(\sqrt{3}) \approx 0.549...$

Yuriy S
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StopReadingThisUsername
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4 Answers4

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You have $$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\sec^2 x}{\sqrt[3]{\tan x}}dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{1}{\sqrt[3]{\tan x}}d\tan x$$

Replacing $t = \tan x$, you have the integration equals $$\int_{1}^{\sqrt{3}} \frac{1}{\sqrt[3]{t}}dt = \frac{3}{2}t^{\frac{2}{3}} \mid_{1}^{\sqrt{3}} = \frac{3}{2}(\sqrt[3]{3}-1)$$

Yuriy S
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GAVD
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Put $u = \tan x\implies I = \displaystyle \int_{1}^{\sqrt{3}} u^{-\frac{1}{3}}du= \left[\frac{3}{2}u^{\frac{2}{3}}\right]|_{u=1}^{u = \sqrt{3}}= \dfrac{3}{2}\left(\sqrt[3]{3}-1\right)$

DeepSea
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$$ \frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{2}{3}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx = \frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{f'(x)}{f(x)}dx$$

This step is wrong since

$$f(x) = (\tan \ x)^{\frac{2}{3}}, \ f'(x) = \frac{2}{3} \cdot (\tan \ x)^{-\frac{1}{3}} \cdot \sec^2x dx$$

Then $$\frac{f'(x)}{f(x)} = \frac{2}{3} \frac{ \sec^2x}{\tan x} \neq \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{2}{3}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx dx$$

Try to use trigonometric identities to simplificate it first and then use substitute:

Let $u = \sqrt[3] {\tan{x}}$, then $u^3 = \tan x $, $2u^2 du = \sec^2 x dx$.

So $$\frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{2}{3}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx = \frac{3}{2}\int_{1}^{ 3^{1/6}}\frac{2}{3}\frac{2u^2}{u} du$$

Yuriy S
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Zau
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Your solution was fine. Just needed to have put $$\frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}f'(x)dx$$ instead of $$\frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{f'(x)}{f(x)}dx$$which makes your solution $$\frac{3}{2}(f(\frac{\pi}{3})-f(\frac{\pi}{4}))$$where f(x) is the function you defined, namely $$f(x) = (\tan \ x)^{\frac{2}{3}}$$This indeed yields the correct answer, $$\frac{3}{2}(3^{\frac{1}{3}}-1)$$

BelowAverageIntelligence
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