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Exercise

∫0π4(πtan3(x))(−πsec3(x))dx\int_0^{\frac{\pi}{4}}\left(\pi tan^3\left(x\right)\right)\left(-\pi sec^3\left(x\right)\right)dx

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Final answer to the exercise

sec⁡(π4)3⋅(−243.9281302123.5754344⋅sec⁡(π4)2+233.525120670.9831249)\sec\left(\frac{\pi }{4}\right)^{3}\cdot \left(-\frac{243.9281302}{123.5754344}\cdot \sec\left(\frac{\pi }{4}\right)^2+\frac{233.5251206}{70.9831249}\right)

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∫0π4 (πtan3(x))(−πsec3(x))dx
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