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∫tan2x⋅sec3x dx\int tan^2x\cdot sec^3x\:dx

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Final answer to the exercise

−58ln⁡∣sec⁡(x)+tan⁡(x)∣−58tan⁡(x)sec⁡(x)+14sec⁡(x)3tan⁡(x)+12ln⁡∣sec⁡(x)+tan⁡(x)∣+tan⁡(x)sec⁡(x)2+C0-\frac{5}{8}\ln\left|\sec\left(x\right)+\tan\left(x\right)\right|-\frac{5}{8}\tan\left(x\right)\sec\left(x\right)+\frac{1}{4}\sec\left(x\right)^3\tan\left(x\right)+\frac{1}{2}\ln\left|\sec\left(x\right)+\tan\left(x\right)\right|+\frac{\tan\left(x\right)\sec\left(x\right)}{2}+C_0

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∫tan2x·sec3x dx
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