∫tan³(θ)sec(θ) dθ + ∫tan⁵(θ)sec(θ) dθ

asked by guest
on Nov 18, 2024 at 11:25 am



You asked:

Evaluate the expression: ∫tan⁡(θ)3sec⁡(θ)dθ+∫tan⁡(θ)5sec⁡(θ)dθ\int{{\tan\left( θ \right)}^{3} \sec\left( θ \right)}dθ + \int{{\tan\left( θ \right)}^{5} \sec\left( θ \right)}dθ

MathBot Answer:

∫tan⁡(θ)3sec⁡(θ)dθ+∫tan⁡(θ)5sec⁡(θ)dθ=C1+C2+1−3cos⁡2(θ)3cos⁡3(θ)−−15cos⁡4(θ)+10cos⁡2(θ)−315cos⁡5(θ)\int{{\tan\left( θ \right)}^{3} \sec\left( θ \right)}dθ + \int{{\tan\left( θ \right)}^{5} \sec\left( θ \right)}dθ = C_{1} + C_{2} + \frac{1 - 3 \cos^{2}{\left(θ \right)}}{3 \cos^{3}{\left(θ \right)}} - \frac{- 15 \cos^{4}{\left(θ \right)} + 10 \cos^{2}{\left(θ \right)} - 3}{15 \cos^{5}{\left(θ \right)}}

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