Question: "Evaluate the following integrals. ∫cos42θdθ"

midtlinjeg

Answered question

2021-10-06

Evaluate the following integrals.

\int \cos^{4} 2 θ d θ

Daphne Broadhurst

Skilled2021-10-07Added 109 answers

Consider the provided integral,

\int \cos^{4} 2 θ d θ

Evaluate the provided integral.
Apply the substitution method,
Let

u = 2 θ \Rightarrow d u = 2 d θ

We get,

\int \cos^{4} 2 θ d θ = \frac{1}{2} \int \cos^{4} u d u

We can write as,

\int \cos^{4} 2 θ d θ = \frac{1}{2} \int \cos^{3} u \cos u d u

\int \cos^{4} 2 θ d θ = \frac{1}{2} \int \cos^{3} u \cos u d u

Now, apply the integration by parts.
Let

u = \cos^{3} u

and

v = \cos u

\int \cos^{4} 2 θ d θ = \frac{1}{2} \int \cos^{3} u \cos u d u

= \frac{1}{2} (\cos^{3} u \sin u - \int - 3 \cos^{2} u \sin u d u) + C

= \frac{1}{2} (\cos^{3} u \sin u + 3 \int \cos^{2} u \sin u d u) + C

= \frac{1}{2} (\cos^{3} u \sin u + 3 \int \frac{1 - \cos 4 u}{8} d u) + C

Simplifying further,

\int \cos^{4} 2 θ d θ = \frac{1}{2} (\cos^{3} u \sin u + \frac{3}{8} \int 1 - \cos 4 u d u) + C

= \frac{1}{2} (\cos^{3} u \sin u + \frac{3}{8} (u - \frac{1}{4} \sin 4 u)) + C

Substitute back

u = 2 θ

we get,

\int \cos^{42} θ d θ = \frac{1}{2} (\cos^{32} θ \sin 2 θ + \frac{3}{8} (2 θ - \frac{1}{4} \sin 8 θ)) + C

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