sec-tan-4-d-

Question Number 167024 by peter frank last updated on 04/Mar/22

\int \sec θ \tan^{4} θ d θ

Answered by cortano1 last updated on 05/Mar/22

let t = \tan θ

I = \int \frac{t^{4}}{\sqrt{1 + t^{2}}} = \int \frac{t^{3} (t)}{\sqrt{1 + t^{2}}} dt

IBP {\begin{cases} u = t^{3} \Rightarrow du = {3 t}^{2} dt \\ v = \int \frac{1}{2} \frac{d (1 + t^{2})}{\sqrt{1 + t^{2}}} = \sqrt{1 + t^{2}} \end{cases}

I = t^{3} \sqrt{1 + t^{2}} - \int {3 t}^{2} \sqrt{1 + t^{2}} dt

I = t^{3} \sqrt{1 + t^{2}} - 3 \int t (t \sqrt{1 + t^{2}}) dt

I = t^{3} \sqrt{1 + t^{2}} - 3 [\frac{1}{3} t \sqrt{{(1 + t^{2})}^{3}} - \frac{1}{3} \int {(1 + t^{2})}^{\frac{3}{2}} dt]

I = t^{3} \sqrt{1 + t^{2}} - t \sqrt{{(1 + t^{2})}^{3}} + \int {(1 + t^{2})}^{\frac{3}{2}} dt

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