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∫π4π323sec4xdx\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}23sec^4xdx

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Final answer to the exercise

23⋅(tan⁡(π3)⋅sec⁡(π3)23+23tan⁡(π3)−(tan⁡(π4)⋅sec⁡(π4)23+23tan⁡(π4)))23\cdot \left(\frac{\tan\left(\frac{\pi }{3}\right)\cdot \sec\left(\frac{\pi }{3}\right)^{2}}{3}+\frac{2}{3}\tan\left(\frac{\pi }{3}\right)- \left(\frac{\tan\left(\frac{\pi }{4}\right)\cdot \sec\left(\frac{\pi }{4}\right)^{2}}{3}+\frac{2}{3}\tan\left(\frac{\pi }{4}\right)\right)\right)

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∫π4 π3 23sec4xdx
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