What is the integral of e^θtan^4(e^θ) ?

The integral of e^θtan^4(e^θ) is 1/3 tan^3(e^θ)+e^θ-tan(e^θ)+C

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integral of e^θtan^4(e^θ)

\int e^{θ} tan^{4} (e^{θ}) d θ

\frac{1}{3} tan^{3} (e^{θ}) + e^{θ} - tan (e^{θ}) + C

Solution steps

\int e^{θ} tan^{4} (e^{θ}) d θ

Apply u-substitution

: \int tan^{4} (u) d u

= \int tan^{4} (u) d u

Apply Integral Reduction

= \frac{tan ^{3} ( u )}{3} - \int tan^{2} (u) d u

\int tan^{2} (u) d u = - u + tan (u)

= \frac{tan ^{3} ( u )}{3} - (- u + tan (u))

Substitute back

u = e^{θ}

= \frac{tan ^{3} ( e ^{θ} )}{3} - (- e^{θ} + tan (e^{θ}))

Simplify

\frac{tan ^{3} ( e ^{θ} )}{3} - (- e^{θ} + tan (e^{θ})) : \frac{1}{3} tan^{3} (e^{θ}) + e^{θ} - tan (e^{θ})

= \frac{1}{3} tan^{3} (e^{θ}) + e^{θ} - tan (e^{θ})

Add a constant to the solution

= \frac{1}{3} tan^{3} (e^{θ}) + e^{θ} - tan (e^{θ}) + C

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What is the integral of e^θtan^4(e^θ) ?
The integral of e^θtan^4(e^θ) is 1/3 tan^3(e^θ)+e^θ-tan(e^θ)+C